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GR 8 Math Powerpoint about Polynomial Techniques

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El documento detalla técnicas de factorización de polinomios, incluyendo trinomios cuadráticos, factores comunes monomiales, diferencia de cuadrados y trinomios cuadrados perfectos. Se explican los pasos necesarios para identificar y aplicar estas técnicas a diferentes tipos de polinomios, así como ejemplos de cada método. Además, se proporcionan ejercicios prácticos y claves de respuesta para mejorar la comprensión.

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FACTORING POLYNOMIALS(TECHNIQUES) By
Quadratic Trinomial I. a=1 pages 17-22 II. a>1 pages 23-28 Factoring Polynomials with Greatest Common Monomial Factor pages 5-8 Difference of 2 Squares pages 9-12 Factoring Perfect Square Trinomials pages 13-16 CONTENTS
Factoring Quadratic Trinoial CONTENTS Sum and Difference of Two Cubes pages 29-32 Factoring by Grouping pages 33-36 Exercises Answer Keys pages 37-38
Factoring is the process of finding the factors. It reverses the multiplication process Factoring
Greatest Common Monomial Factor The polynomial of the highest degree that is an exact divisor of each of two polynomials. The polynomial of the highest degree that is an exact divisor of each of two polynomials.
How do we factor polynomials using the Greatest Common Monomial Factor Method? Given example: 20x²+12x Find the GCF with the lowest exponent of all the terms in a polynomial. 1. GCMF: 4X 2. Factor out the GCF from each terms. 4x(5x)+4x(3) 3. Use distributive property to factor and simplify expressions Final Answer: 4x(5x+3) NOTE: Remember that distributive property says that a(b+c) is = (a)(b)+(a)(c) Greatest Common Monomial Factor
Greatest Common Monomial Factor Examples 9x⁹+54x⁵ 5x⁷+30x⁴-10x³ x²y⁴+xy⁶z²+x⁵y² 9x⁹+54x⁵ 1. Find the GCF of Each Term factors of: 9x⁹: (9)(1)(x)(x)(x)(x)(x)(x)(x)(x)(x) 54x⁵ : (9)(6)(x)(x)(x)(x)(x) The GCF is: 9x⁵ 2. Use the GCF to factor each terms. 9x⁹+54x⁵ = 9x⁵(x⁴)+9x⁵ (6) 3. Make use of the Distributive Property: The final answer is 9x⁵ (x⁴+6) 5x⁷+30x⁴-10x³ 1. Find the GCF of Each Term factors of: 5x⁷: (1)(5)(x)(x)(x)(x)(x)(x)(x) 30x⁴: (5)(6)(x)(x)(x)(x) 10x³: (5)(2)(x)(x)(x) The GCF is: 5x³ 2. Use the GCF to factor each terms. 5x⁷+30x⁴-10x³ = 5x³ (x⁴)+ 5x³(6x)-5x³ (2) 3. Make use of the Distributive Property: The final answer is 5x³ (x⁴ +6x-2) x² y⁴ +xy⁶z² +x⁵ y² 1. Find the GCF of Each Term factors of: x² y⁴: (x)(x)(y)(y) xy⁶z² : (x)(y)(y)(y)(y)(y)(y)(z)(z) x⁵ y² :(x)(x)(x)(x)(x)(y)(y) The GCF is: xy² 2. Use the GCF to factor each terms. x²y⁴+xy⁶z²+x⁵y²= xy²(xy²)+xy² (y⁴z²) +xy² (x⁴) 3. Make use of the Distributive Property: The final answer is xy²(xy² +y⁴z² +x⁴)
Try These. 1. 15x²-45x⁶-60x⁸ 2. 88x⁵+11y² 3. abc-ab⁵d³+ab⁹c⁴
Difference of Two Squares In mathematics, the difference of two squares is a squared (multiplied by itself) number subtracted from another squared number. Every difference of squares may be factored according to the identity. in elementary algebra. -Wiki
How do we factor polynomials using Difference of Two Squares? The two terms must be a polynomial. There must be a minus sign between the two terms Conditions for a polynomial to be DOTS: 1. 2. The two terms must be both squares. There must be a minus sign between the two terms Conditions for a polynomial to be DOTS: 1. 2. Finding the square root of each terms. Write each terms’ factorization as sum and difference of the square root. We can factor a polynomial using DoTS by: 1. 2. Finding the square root of each terms. Write each terms’ factorization as sum and difference of the square root. We can factor a polynomial using DoTS by: 1. 2. Difference of Two Squares
x²- 81 9x⁶ -25y⁴ Finding the square root of each terms x²-81 1. x²: (x)² 81: (9)² 2. Write each terms’ factorization as sum and difference of the square root. Final Answer: (x+9)(x-9) Finding the square root of each terms 9x⁶-25y⁴ 1. 9x⁶: (3)² (x³)² 25y⁴: (5)² (y²)² 2. Write each terms’ factorization as sum and difference of the square root. Final Answer: (3x³+5y²)(3x³-5y²) 1-25q⁴ Finding the square root of each terms 1-25q⁴ 1. 1: (1)² 25q⁴: (5q²)(²) 2. Write each terms’ factorization as sum and difference of the square root. Final Answer: (1+5q²)(1-5q² ) Difference of Two Squares
Try These. 1. 144h²-49p² 2. a²b ²– 16 3. 100g²-25
Perfect Square Trinomials In mathematics, Perfect Square Trinomial is an expression obtained from the square of binomial equation. The perfect square formula takes the following: (ax)²+2abx+b² = (ax+b)²
Identify the square root of the first and last terms. Determine the operation of the middle term to be used on How do we factor? 1. 2. separating the factored form. 3. Write the final answer as square of binomial. x²+2xy+y²=(x+y)² Identify the square root of the first and last terms. Determine the operation of the middle term to be used on How do we factor? 1. 2. separating the factored form. 3. Write the final answer as square of binomial. x²+2xy+y²=(x+y)² Perfect Square Trinomials The first and last term must be perfect squares. There must be no minus sign before the first and last term. If you multiply its first and last term, the result must be its additive inverse. Conditions for a polynomial to be a PST: 1. 2. 3. The first and last term must be perfect squares. There must be no minus sign before the first and last term. If you multiply its first and last term, the result must be its additive inverse. Conditions for a polynomial to be a PST: 1. 2. 3. How do we factor polynomials using Perfect Square Trinomials?
Identify the square root of the first and last terms. 1. x²: (x)² 4: (2) ² 2. Determine the operation of the middle term to be used on separating the factored form. x² +4x+4 3. Write the final answer as square of binomial. (x+2)² x²+4x+4 x²+4x+4 16x² +72x+81 16x² +72x+81 Identify the square root of the first and last terms. 1. 16x ²: (4x)² 81: (9)² 2. Determine the operation of the middle term to be used on separating the factored form. 16x² +72x+81 3. Write the final answer as square of binomial. (4x +9)² 25p²-60pq+36q² 25p²-60pq+36q² Identify the square root of the first and last terms. 1. 25p²: (5p)² 36q²: (6q)² 2. Determine the operation of the middle term to be used on separating the factored form. 25p² -60pq+36q² 3. Write the final answer as square of binomial. (5p-6q)²
Try These. 1. r²+6r+9 2. 25+40x+16x² 3. (a+b)²+10(a+b)+25
Quadratic Trinomial (a=1) A quadratic trinomial is a trinomial of the form ax² +bx+c, where a=1.
List all pairs of integers whose product is equal to the last term Choose a pair, m and n, whose sum is the middle term: m+n=b Write the factorization as x² +bx+c=(x+m)(x+n). If the terms of trinomial do not have a factor, then terms of binomial factor cannot have a common factor If constant term of a trinomial is: Factoring x² + bx +c 1. 2. 3. Points to remember in factoring Quadratic Trinomial. 1. 2. a. positive, the constant term of the binomials have the same signs as the coeffecient of x in the trinomial. b. negative, the constant terms of the binomials have opposite signs How do we factor polynomials using Quadratic Trinomials (a=1)
Quadratic Trinomial (a=1) Examples List all pairs of integers whose product is equal to the last term 1. last term is 10: Factors of 10: 5 and 2, 10 and 1 2. Choose a pair, m and n, whose sum is the middle term: m+n=b The factors 5 and 2 is equal to the middle term 7. 3. Write the factorization as x² +bx+c=(x+m)(x+n) Final answer is (x+5)(x+2) 1.Example: x² +7x+10 Some Factors of 10 equal to 7 -5 and -2 5 and 2 -1 and -10 1 and 10 -7 7 -11 11
Quadratic Trinomial (a=1) Examples List all pairs of integers whose product is equal to the last term 1. last term is 14 Factors of 14: 1,2,7,14 2. Choose a pair, m and n, whose sum is the middle term: m+n=b The factors -2 and -7 is equal to the middle term -9 3. Write the factorization as x² +bx+c=(x+m)(x+n) Final answer is (x-2)(x-7) 2. Example: x²-9x+14 Some Factors of 14 equal to -9x -1 and -14 1 and 14 2 and 7 -2 and -7 -7 7 -9 9
Quadratic Trinomial (a=1) Examples List all pairs of integers whose product is equal to the last term 1. last term is 18 Factors of 14: 1,2,3,6, 9,18 2. Choose a pair, m and n, whose sum is the middle term: m+n=b The factors -2 and -9 is equal to the middle term -11 3. Write the factorization as x² +bx+c=(x+m)(x+n) Final answer is (x-2)(x-9) 3. x²-11x+18 Some Factors of 18 equal to -11x 1,18 -1,-18 -2,-9 2,9 -3, -6 3, 6 19 -19 -11 11 -9 9
1. p²+14p+40 2. b²-8b+15 3.r²+r-20 Try These. 1. p²+14p+40 2. b²-8b+15 3.r²+r-20
Quadratic Trinomial (a>1) A Quadratic Trinomial (a>1) is a trinomial of the form ax² +bx+c, where a>1
How do we factor polynomials using Quadratic Trinomials (a=1) Start by finding two numbers that multiply to ac and b Use these numbers to split up the x-term. Use grouping to factor the quadratic expression. In general, we can use the following steps to factor a quadratic form: ax²+ bx+c 1. 2. 3. 4. Points to remember in factoring Quadratic Trinomial. If the terms of trinomial do not have a factor, then terms of binomial factor cannot have a common factor If constant term of a trinomial is: a. positive, the constant term of the binomials have the same signs as the coeffecient of x in the trinomial. b. negative, the constant terms of the binomials have opposite signs
Quadratic Trinomial (a>1) Examples Factors of 56 whose sum is 15 1, 56 4, 14 7,8 57 30 18 15 Start by finding two numbers that multiply to ac and add to b. In general, we can use the following steps to factor a quadratic form 2x² +15x+28 1. (2)(28)= 56 Factors of 56 that adds to 15: 7, 8 2. Use these numbers to split up the x-term. 2x² +15x+28=2x²+(7x+8x)+28 3. Use grouping to factor the quadratic equation 2x² +7x+8x+28 =(2x² +7x)+(8x+28) = (2x+7)(x+4) is the final answer. 1.Example: 2x+15x+28 2, 28
Quadratic Trinomial (a>1) Examples Factors of -24 whose sum is -10 Start by finding two numbers that multiply to ac and add to b. In general, we can use the following steps to factor a quadratic form 3x² -10x-8 1. (3)(-8)=-24 Factors of -24 that adds to -10: -12 and 2 2. Use these numbers to split up the x-term. 3x²-10x-8 = 3x²-12+2-8 3. Use grouping to factor the quadratic equation 3x² +2x-12x-8= x(3x+2)-4(3x+2)= (3x+2)(x-4) is the final answer. 2. Example:3x² -10x-8 -1,24 1,-24 -2,12 -12,2 -3,8 3,-8 -4,6 4,-6 23 -23 10 -10 5 -5 2 -2
Quadratic Trinomial (a>1) Examples Factors of 24 whose sum is -25 -1, -24 -3, -8 -4, -6 -25 -11 -10 Start by finding two numbers that multiply to ac and add to b. In general, we can use the following steps to factor a quadratic form 6x²-25x+4 1. (6)(4)=24 Factors of 24 that adds to -25: (1)(24) 2. Use these numbers to split up the x-term. 6x²-25x+4= 6x²-x-24x+4 3. Use grouping to factor the quadratic equation 6x²-x-24+4 =x(6x-1)-4(6x-1) (6x-1)(x-4) = (6x-1)(x-4) is the final answer. 3.Example:6x² -25x+4 -2, -12 -14
Try These. 1. 12x²-29x+15 2. 7a²+53a+28 3. 6x² +37x+6
Sum and Difference of two Cubes A binomial with each term being cubed and it has either addition od subtraction as the middle sign
a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cuberoots. c. square the last term of the binomial to be put on the last term of the trinomial. points to remember in factoring sum and difference of two cubes: a. The sign separating the trinomials’ middle and last term must be always addition or positive. b. The sign separating the trinomials’ first and middle term must be opposite to the given sign How do we factor polynomials using the sum and difference of two cubes? Sum and Difference of two Cubes
Sum and Difference of two Cubes a+b³ a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. a+b³ =(a+b) 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cube roots. square the last term of the binomial to be put on the last term of the trinomial. 1a. binomial factor: (a+b) 2a. trinomial factor’ first term: (a)² = a² b. Middle term: (a)(b)=ab b. last term: (b)² =b² Combine : (a+b)(a²-ab+b²) Hence, the final answer is a+b³ = (a+b)(a²-ab+b²) a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. y³-8=( y-2) 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cube roots. c. square the last term of the binomial to be put on the last term of the trinomial. 1a. binomial factor: (y-2) 2a. trinomial factor’ first term: (y)² b. Middle term: (2)(y) =2y c. last term: (2)² =4 Combine : (y-2)(y² +2y+4) Hence, the final answer is y³-8=(y-2)(y² +2y+4) y³-8 a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. 27x³ +64y³ =(3x+4y) 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cube roots. c. square the last term of the binomial to be put on the last term of the trinomial. 1.a. binomial factor: (3x+4y) 2. a. trinomial factor’ first term: (3x)² =9x² b. Middle term: (3x)(4y)=12xy c. last term: (4y)² =16y² Combine : (3x+4y)(9x²-12xy+16y²) Hence, the final answer is 27x³ +64y³ = (3x+4y)(9x²-12xy+16y²) 27x³ +64y³
Try These. 1. 27m³-125 2. 500x³ +256 3. 375-81a³
Factoring by Grouping Just like it says, factoring by grouping means that you will group terms with common factors before factoring.
Step 1: Group terms with Common Monomial Factor. Step 2: Factor out the GCF from each separate groups. Step 3: Factor out the common binomial factor. How do we factor polynomials using Factoring by Grouping? Factoring by Grouping
2x² +8x+3x+12 Step 1: Group terms with Common Monomial Factor. (2x² +8x)+(3x+12) Step 2: Factor out the GCF from each separate groups. 2x(x+4)+3(x+4) Step 3: Factor out the common binomial factor. (2x+3)(x+4) 3x²-6x-4x+8 Step 1: Group terms with Common Monomial Factor. (3x²-6x)+(-4x+8) Step 2: Factor out the GCF from each separate groups. 3x(x-2)-4(x-2) Step 3: Factor out the common binomial factor. (x-2)(3x-4) 2x³+3x-10x²-15 Step 1: Group terms with Common Monomial Factor. (2x³-10x²)+(3x-15) Step 2: Factor out the GCF from each separate groups. 2x² (x-5)+3(x-5) Step 3: Factor out the common binomial factor. (x-5)(2x² +3) Factoring by Grouping
Try These. 1. 5mn + 25m + 3n³ + 15n² 2. 4b³+ b² + 8b + 2 3. 28k³ − 4k²− 35k + 5
Answer Keys
15x² (1-3x⁴-4x⁶) Greatest Common Monomial Factor 1. 2. 11(8x⁵ +y) 3. ab(c-b⁴d³+b⁸c⁴ DoTS 1. (12h+7p)(12h-7p) 2.(ab+4)(ab-4) 3. 25(2g+1)(2g-1) Perfect Square Trinomial 1.(r+3)² 2.(5+4x)² 3. (a+b+5)² Quadratic Trinomial a=1 1. (p+4)(p+10) 2. (b-5)(b-3) 3. (r-4)(r-5) Quadratic Trinomial a>1 1. (4x-3)(3x-5) 2. (a+7)(7a+4) 3. (6x+1)(x+6) Sum and Dif. of two cubes 1. (3m-5)(9m² +15m+25) 2. 4(5x+4)(25x²-20x+16) 3. 3(5-3a)(25+15a+9a²) Factor by Grouping 1.(5m+3n²)(n+5) 2.(4b+1)(b² +2) 3. (7k-1)(4k²-5)
-Oronce,O.A.,&Mendoza,M.O.(2019).E-Math: WorktextinMathematics.REXBookStore. -KhanAcademy.AccessedJanuary31,2020 https://www.khanacademy.org. -Factoring.(2019,February22).InWikipedia.URL -OnlineLearningMath.com.AccessedJanuary31,2020. https://www.onlinemathlearning.com/factoring-by- grouping.html Resources

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GR 8 Math Powerpoint about Polynomial Techniques

  • 2. Quadratic Trinomial I. a=1 pages 17-22 II. a>1 pages 23-28 Factoring Polynomials with Greatest Common Monomial Factor pages 5-8 Difference of 2 Squares pages 9-12 Factoring Perfect Square Trinomials pages 13-16 CONTENTS
  • 3. Factoring Quadratic Trinoial CONTENTS Sum and Difference of Two Cubes pages 29-32 Factoring by Grouping pages 33-36 Exercises Answer Keys pages 37-38
  • 4. Factoring is the process of finding the factors. It reverses the multiplication process Factoring
  • 5. Greatest Common Monomial Factor The polynomial of the highest degree that is an exact divisor of each of two polynomials. The polynomial of the highest degree that is an exact divisor of each of two polynomials.
  • 6. How do we factor polynomials using the Greatest Common Monomial Factor Method? Given example: 20x²+12x Find the GCF with the lowest exponent of all the terms in a polynomial. 1. GCMF: 4X 2. Factor out the GCF from each terms. 4x(5x)+4x(3) 3. Use distributive property to factor and simplify expressions Final Answer: 4x(5x+3) NOTE: Remember that distributive property says that a(b+c) is = (a)(b)+(a)(c) Greatest Common Monomial Factor
  • 7. Greatest Common Monomial Factor Examples 9x⁹+54x⁵ 5x⁷+30x⁴-10x³ x²y⁴+xy⁶z²+x⁵y² 9x⁹+54x⁵ 1. Find the GCF of Each Term factors of: 9x⁹: (9)(1)(x)(x)(x)(x)(x)(x)(x)(x)(x) 54x⁵ : (9)(6)(x)(x)(x)(x)(x) The GCF is: 9x⁵ 2. Use the GCF to factor each terms. 9x⁹+54x⁵ = 9x⁵(x⁴)+9x⁵ (6) 3. Make use of the Distributive Property: The final answer is 9x⁵ (x⁴+6) 5x⁷+30x⁴-10x³ 1. Find the GCF of Each Term factors of: 5x⁷: (1)(5)(x)(x)(x)(x)(x)(x)(x) 30x⁴: (5)(6)(x)(x)(x)(x) 10x³: (5)(2)(x)(x)(x) The GCF is: 5x³ 2. Use the GCF to factor each terms. 5x⁷+30x⁴-10x³ = 5x³ (x⁴)+ 5x³(6x)-5x³ (2) 3. Make use of the Distributive Property: The final answer is 5x³ (x⁴ +6x-2) x² y⁴ +xy⁶z² +x⁵ y² 1. Find the GCF of Each Term factors of: x² y⁴: (x)(x)(y)(y) xy⁶z² : (x)(y)(y)(y)(y)(y)(y)(z)(z) x⁵ y² :(x)(x)(x)(x)(x)(y)(y) The GCF is: xy² 2. Use the GCF to factor each terms. x²y⁴+xy⁶z²+x⁵y²= xy²(xy²)+xy² (y⁴z²) +xy² (x⁴) 3. Make use of the Distributive Property: The final answer is xy²(xy² +y⁴z² +x⁴)
  • 8. Try These. 1. 15x²-45x⁶-60x⁸ 2. 88x⁵+11y² 3. abc-ab⁵d³+ab⁹c⁴
  • 9. Difference of Two Squares In mathematics, the difference of two squares is a squared (multiplied by itself) number subtracted from another squared number. Every difference of squares may be factored according to the identity. in elementary algebra. -Wiki
  • 10. How do we factor polynomials using Difference of Two Squares? The two terms must be a polynomial. There must be a minus sign between the two terms Conditions for a polynomial to be DOTS: 1. 2. The two terms must be both squares. There must be a minus sign between the two terms Conditions for a polynomial to be DOTS: 1. 2. Finding the square root of each terms. Write each terms’ factorization as sum and difference of the square root. We can factor a polynomial using DoTS by: 1. 2. Finding the square root of each terms. Write each terms’ factorization as sum and difference of the square root. We can factor a polynomial using DoTS by: 1. 2. Difference of Two Squares
  • 11. x²- 81 9x⁶ -25y⁴ Finding the square root of each terms x²-81 1. x²: (x)² 81: (9)² 2. Write each terms’ factorization as sum and difference of the square root. Final Answer: (x+9)(x-9) Finding the square root of each terms 9x⁶-25y⁴ 1. 9x⁶: (3)² (x³)² 25y⁴: (5)² (y²)² 2. Write each terms’ factorization as sum and difference of the square root. Final Answer: (3x³+5y²)(3x³-5y²) 1-25q⁴ Finding the square root of each terms 1-25q⁴ 1. 1: (1)² 25q⁴: (5q²)(²) 2. Write each terms’ factorization as sum and difference of the square root. Final Answer: (1+5q²)(1-5q² ) Difference of Two Squares
  • 12. Try These. 1. 144h²-49p² 2. a²b ²– 16 3. 100g²-25
  • 13. Perfect Square Trinomials In mathematics, Perfect Square Trinomial is an expression obtained from the square of binomial equation. The perfect square formula takes the following: (ax)²+2abx+b² = (ax+b)²
  • 14. Identify the square root of the first and last terms. Determine the operation of the middle term to be used on How do we factor? 1. 2. separating the factored form. 3. Write the final answer as square of binomial. x²+2xy+y²=(x+y)² Identify the square root of the first and last terms. Determine the operation of the middle term to be used on How do we factor? 1. 2. separating the factored form. 3. Write the final answer as square of binomial. x²+2xy+y²=(x+y)² Perfect Square Trinomials The first and last term must be perfect squares. There must be no minus sign before the first and last term. If you multiply its first and last term, the result must be its additive inverse. Conditions for a polynomial to be a PST: 1. 2. 3. The first and last term must be perfect squares. There must be no minus sign before the first and last term. If you multiply its first and last term, the result must be its additive inverse. Conditions for a polynomial to be a PST: 1. 2. 3. How do we factor polynomials using Perfect Square Trinomials?
  • 15. Identify the square root of the first and last terms. 1. x²: (x)² 4: (2) ² 2. Determine the operation of the middle term to be used on separating the factored form. x² +4x+4 3. Write the final answer as square of binomial. (x+2)² x²+4x+4 x²+4x+4 16x² +72x+81 16x² +72x+81 Identify the square root of the first and last terms. 1. 16x ²: (4x)² 81: (9)² 2. Determine the operation of the middle term to be used on separating the factored form. 16x² +72x+81 3. Write the final answer as square of binomial. (4x +9)² 25p²-60pq+36q² 25p²-60pq+36q² Identify the square root of the first and last terms. 1. 25p²: (5p)² 36q²: (6q)² 2. Determine the operation of the middle term to be used on separating the factored form. 25p² -60pq+36q² 3. Write the final answer as square of binomial. (5p-6q)²
  • 16. Try These. 1. r²+6r+9 2. 25+40x+16x² 3. (a+b)²+10(a+b)+25
  • 17. Quadratic Trinomial (a=1) A quadratic trinomial is a trinomial of the form ax² +bx+c, where a=1.
  • 18. List all pairs of integers whose product is equal to the last term Choose a pair, m and n, whose sum is the middle term: m+n=b Write the factorization as x² +bx+c=(x+m)(x+n). If the terms of trinomial do not have a factor, then terms of binomial factor cannot have a common factor If constant term of a trinomial is: Factoring x² + bx +c 1. 2. 3. Points to remember in factoring Quadratic Trinomial. 1. 2. a. positive, the constant term of the binomials have the same signs as the coeffecient of x in the trinomial. b. negative, the constant terms of the binomials have opposite signs How do we factor polynomials using Quadratic Trinomials (a=1)
  • 19. Quadratic Trinomial (a=1) Examples List all pairs of integers whose product is equal to the last term 1. last term is 10: Factors of 10: 5 and 2, 10 and 1 2. Choose a pair, m and n, whose sum is the middle term: m+n=b The factors 5 and 2 is equal to the middle term 7. 3. Write the factorization as x² +bx+c=(x+m)(x+n) Final answer is (x+5)(x+2) 1.Example: x² +7x+10 Some Factors of 10 equal to 7 -5 and -2 5 and 2 -1 and -10 1 and 10 -7 7 -11 11
  • 20. Quadratic Trinomial (a=1) Examples List all pairs of integers whose product is equal to the last term 1. last term is 14 Factors of 14: 1,2,7,14 2. Choose a pair, m and n, whose sum is the middle term: m+n=b The factors -2 and -7 is equal to the middle term -9 3. Write the factorization as x² +bx+c=(x+m)(x+n) Final answer is (x-2)(x-7) 2. Example: x²-9x+14 Some Factors of 14 equal to -9x -1 and -14 1 and 14 2 and 7 -2 and -7 -7 7 -9 9
  • 21. Quadratic Trinomial (a=1) Examples List all pairs of integers whose product is equal to the last term 1. last term is 18 Factors of 14: 1,2,3,6, 9,18 2. Choose a pair, m and n, whose sum is the middle term: m+n=b The factors -2 and -9 is equal to the middle term -11 3. Write the factorization as x² +bx+c=(x+m)(x+n) Final answer is (x-2)(x-9) 3. x²-11x+18 Some Factors of 18 equal to -11x 1,18 -1,-18 -2,-9 2,9 -3, -6 3, 6 19 -19 -11 11 -9 9
  • 22. 1. p²+14p+40 2. b²-8b+15 3.r²+r-20 Try These. 1. p²+14p+40 2. b²-8b+15 3.r²+r-20
  • 23. Quadratic Trinomial (a>1) A Quadratic Trinomial (a>1) is a trinomial of the form ax² +bx+c, where a>1
  • 24. How do we factor polynomials using Quadratic Trinomials (a=1) Start by finding two numbers that multiply to ac and b Use these numbers to split up the x-term. Use grouping to factor the quadratic expression. In general, we can use the following steps to factor a quadratic form: ax²+ bx+c 1. 2. 3. 4. Points to remember in factoring Quadratic Trinomial. If the terms of trinomial do not have a factor, then terms of binomial factor cannot have a common factor If constant term of a trinomial is: a. positive, the constant term of the binomials have the same signs as the coeffecient of x in the trinomial. b. negative, the constant terms of the binomials have opposite signs
  • 25. Quadratic Trinomial (a>1) Examples Factors of 56 whose sum is 15 1, 56 4, 14 7,8 57 30 18 15 Start by finding two numbers that multiply to ac and add to b. In general, we can use the following steps to factor a quadratic form 2x² +15x+28 1. (2)(28)= 56 Factors of 56 that adds to 15: 7, 8 2. Use these numbers to split up the x-term. 2x² +15x+28=2x²+(7x+8x)+28 3. Use grouping to factor the quadratic equation 2x² +7x+8x+28 =(2x² +7x)+(8x+28) = (2x+7)(x+4) is the final answer. 1.Example: 2x+15x+28 2, 28
  • 26. Quadratic Trinomial (a>1) Examples Factors of -24 whose sum is -10 Start by finding two numbers that multiply to ac and add to b. In general, we can use the following steps to factor a quadratic form 3x² -10x-8 1. (3)(-8)=-24 Factors of -24 that adds to -10: -12 and 2 2. Use these numbers to split up the x-term. 3x²-10x-8 = 3x²-12+2-8 3. Use grouping to factor the quadratic equation 3x² +2x-12x-8= x(3x+2)-4(3x+2)= (3x+2)(x-4) is the final answer. 2. Example:3x² -10x-8 -1,24 1,-24 -2,12 -12,2 -3,8 3,-8 -4,6 4,-6 23 -23 10 -10 5 -5 2 -2
  • 27. Quadratic Trinomial (a>1) Examples Factors of 24 whose sum is -25 -1, -24 -3, -8 -4, -6 -25 -11 -10 Start by finding two numbers that multiply to ac and add to b. In general, we can use the following steps to factor a quadratic form 6x²-25x+4 1. (6)(4)=24 Factors of 24 that adds to -25: (1)(24) 2. Use these numbers to split up the x-term. 6x²-25x+4= 6x²-x-24x+4 3. Use grouping to factor the quadratic equation 6x²-x-24+4 =x(6x-1)-4(6x-1) (6x-1)(x-4) = (6x-1)(x-4) is the final answer. 3.Example:6x² -25x+4 -2, -12 -14
  • 28. Try These. 1. 12x²-29x+15 2. 7a²+53a+28 3. 6x² +37x+6
  • 29. Sum and Difference of two Cubes A binomial with each term being cubed and it has either addition od subtraction as the middle sign
  • 30. a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cuberoots. c. square the last term of the binomial to be put on the last term of the trinomial. points to remember in factoring sum and difference of two cubes: a. The sign separating the trinomials’ middle and last term must be always addition or positive. b. The sign separating the trinomials’ first and middle term must be opposite to the given sign How do we factor polynomials using the sum and difference of two cubes? Sum and Difference of two Cubes
  • 31. Sum and Difference of two Cubes a+b³ a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. a+b³ =(a+b) 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cube roots. square the last term of the binomial to be put on the last term of the trinomial. 1a. binomial factor: (a+b) 2a. trinomial factor’ first term: (a)² = a² b. Middle term: (a)(b)=ab b. last term: (b)² =b² Combine : (a+b)(a²-ab+b²) Hence, the final answer is a+b³ = (a+b)(a²-ab+b²) a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. y³-8=( y-2) 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cube roots. c. square the last term of the binomial to be put on the last term of the trinomial. 1a. binomial factor: (y-2) 2a. trinomial factor’ first term: (y)² b. Middle term: (2)(y) =2y c. last term: (2)² =4 Combine : (y-2)(y² +2y+4) Hence, the final answer is y³-8=(y-2)(y² +2y+4) y³-8 a. For the binomial factor, Find the cube root of the first and last term, copy the sign of the given. 1. 27x³ +64y³ =(3x+4y) 2. a. For the trinomial factor, square the first term of the binomial factor to be put on the first term of the trinomial factor. b. For the middle term, multiply the cube roots. c. square the last term of the binomial to be put on the last term of the trinomial. 1.a. binomial factor: (3x+4y) 2. a. trinomial factor’ first term: (3x)² =9x² b. Middle term: (3x)(4y)=12xy c. last term: (4y)² =16y² Combine : (3x+4y)(9x²-12xy+16y²) Hence, the final answer is 27x³ +64y³ = (3x+4y)(9x²-12xy+16y²) 27x³ +64y³
  • 32. Try These. 1. 27m³-125 2. 500x³ +256 3. 375-81a³
  • 33. Factoring by Grouping Just like it says, factoring by grouping means that you will group terms with common factors before factoring.
  • 34. Step 1: Group terms with Common Monomial Factor. Step 2: Factor out the GCF from each separate groups. Step 3: Factor out the common binomial factor. How do we factor polynomials using Factoring by Grouping? Factoring by Grouping
  • 35. 2x² +8x+3x+12 Step 1: Group terms with Common Monomial Factor. (2x² +8x)+(3x+12) Step 2: Factor out the GCF from each separate groups. 2x(x+4)+3(x+4) Step 3: Factor out the common binomial factor. (2x+3)(x+4) 3x²-6x-4x+8 Step 1: Group terms with Common Monomial Factor. (3x²-6x)+(-4x+8) Step 2: Factor out the GCF from each separate groups. 3x(x-2)-4(x-2) Step 3: Factor out the common binomial factor. (x-2)(3x-4) 2x³+3x-10x²-15 Step 1: Group terms with Common Monomial Factor. (2x³-10x²)+(3x-15) Step 2: Factor out the GCF from each separate groups. 2x² (x-5)+3(x-5) Step 3: Factor out the common binomial factor. (x-5)(2x² +3) Factoring by Grouping
  • 36. Try These. 1. 5mn + 25m + 3n³ + 15n² 2. 4b³+ b² + 8b + 2 3. 28k³ − 4k²− 35k + 5
  • 38. 15x² (1-3x⁴-4x⁶) Greatest Common Monomial Factor 1. 2. 11(8x⁵ +y) 3. ab(c-b⁴d³+b⁸c⁴ DoTS 1. (12h+7p)(12h-7p) 2.(ab+4)(ab-4) 3. 25(2g+1)(2g-1) Perfect Square Trinomial 1.(r+3)² 2.(5+4x)² 3. (a+b+5)² Quadratic Trinomial a=1 1. (p+4)(p+10) 2. (b-5)(b-3) 3. (r-4)(r-5) Quadratic Trinomial a>1 1. (4x-3)(3x-5) 2. (a+7)(7a+4) 3. (6x+1)(x+6) Sum and Dif. of two cubes 1. (3m-5)(9m² +15m+25) 2. 4(5x+4)(25x²-20x+16) 3. 3(5-3a)(25+15a+9a²) Factor by Grouping 1.(5m+3n²)(n+5) 2.(4b+1)(b² +2) 3. (7k-1)(4k²-5)