Factoring Polynomials

Greatest Common Factor The simplest method of factoring a polynomial is to factor out the greatest common factor (GCF) of each term. Example: Factor 18x3 + 60x. 18x3 = 2 · 3 · 3 · x · x · x Factor each term. = (2 · 3 · x) · 3 · x · x 60x = 2 · 2 · 3 · 5 · x = (2 · 3 · x) · 2 · 5 GCF = 6x Find the GCF. 18x3 + 60x = 6x (3x2) + 6x (10) Apply the distributive law to factor the polynomial. = 6x (3x2 + 10) Check the answer by multiplication. 6x (3x2 + 10) = 6x (3x2) + 6x (10) = 18x3 + 60x Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Greatest Common Factor

A common binomial factor can be factored out of certain expressions. Example: Factor 4x2 – 12x + 20. GCF = 4. = 4(x2 – 3x + 5) Check the answer. 4(x2 – 3x + 5) = 4x2 – 12x + 20 A common binomial factor can be factored out of certain expressions. Example: Factor the expression 5(x + 1) – y(x + 1). 5(x + 1) – y(x + 1) = (x + 1) (5 – y) (x + 1) (5 – y) = 5(x + 1) – y(x + 1) Check. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

A difference of two squares can be factored using the formula a2 – b2 = (a + b)(a – b). Example: Factor x2 – 9y2. x2 – 9y2 = (x)2 – (3y)2 Write terms as perfect squares. = (x + 3y)(x – 3y) Use the formula. The same method can be used to factor any expression which can be written as a difference of squares. Example: Factor (x + 1)2 – 25y 4. (x + 1)2 – 25y 4 = (x + 1)2 – (5y2)2 = [(x + 1) + (5y2)][(x + 1) – (5y2)] = (x + 1 + 5y2)(x + 1 – 5y2) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Difference of Squares

Examples: 1. Factor 2xy + 3y – 4x – 6. Some polynomials can be factored by grouping terms to produce a common binomial factor. Examples: 1. Factor 2xy + 3y – 4x – 6. Notice the sign! 2xy + 3y – 4x – 6 = (2xy + 3y) – (4x + 6) Group terms. = y (2x + 3) – 2(2x + 3) Factor each pair of terms. = (2x + 3) ( y – 2) Factor out the common binomial. 2. Factor 2a2 + 3bc – 2ab – 3ac. 2a2 + 3bc – 2ab – 3ac = 2a2 – 2ab + 3bc – 3ac Rearrange terms. = (2a2 – 2ab) + (3bc – 3ac) Group terms. = 2a(a – b) + 3c(b – a) Factor. = 2a(a – b) – 3c(a – b) b – a = – (a – b). = (a – b) (2a – 3c) Factor. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Examples: Factor

Therefore, x2 + 3x + 2 = (x + 1)(x + 2). To factor a simple trinomial of the form x2 + bx + c, express the trinomial as the product of two binomials. For example, x2 + 10x + 24 = (x + 4)(x + 6). Factoring these trinomials is based on reversing the distributive property process. Example: Factor x2 + 3x + 2. Express the trinomial as a product of two binomials with leading term x and unknown constant terms a and b. x2 + 3x + 2 = (x + a)(x + b) = x2 + bx + ax + ba Multiply the binomials. = x2 + (b + a) x + ba Since ab = 2 and a + b = 3, it follows that a = 1 and b = 2. = x2 + (1 + 2) x + 1 · 2 (Product-sum method) Therefore, x2 + 3x + 2 = (x + 1)(x + 2). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Factor x2 + bx + c

It follows that both a and b are negative. Example: Factor x2 – 8x + 15. x2 – 8x + 15 = (x + a)(x + b) = x2 + (a + b)x + ab Therefore a + b = -8 and ab = 15. It follows that both a and b are negative. Sum Negative Factors of 15 - 1, - 15 -15 -3, - 5 - 8 x2 – 8x + 15 = (x – 3)(x – 5). Check: = x2 – 8x + 15. (x – 3)(x – 5) = x2 – 5x – 3x + 15 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

two positive factors of 36 13 36 Example: Factor x2 + 13x + 36. x2 + 13x + 36 = (x + a)(x + b) = x2 + (a + b) x + ab Therefore a and b are: two positive factors of 36 Sum Positive Factors of 36 whose sum is 13. 1, 36 37 2, 18 20 15 3, 12 4, 9 13 6, 6 12 = (x + 4)(x + 9) x2 + 13x + 36 Check: (x + 4)(x + 9) = x2 + 9x + 4x + 36 = x2 + 13x + 36. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

Example: Factor Completely A polynomial is factored completely when it is written as a product of factors that can not be factored further. Example: Factor 4x3 – 40x2 + 100x. 4x3 – 40x2 + 100x The GCF is 4x. = 4x(x2 – 10x + 25) Use distributive property to factor out the GCF. = 4x(x – 5)(x – 5) Factor the trinomial. Check: 4x(x – 5)(x – 5) = 4x(x2 – 5x – 5x + 25) = 4x(x2 – 10x + 25) = 4x3 – 40x2 + 100x Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor Completely

Factoring Polynomials of the Form ax2 + bx + c Factoring complex trinomials of the form ax2 + bx + c, (a  1) can be done by decomposition or cross-check method. Example: Factor 3x2 + 8x + 4. 3  4 = 12 Decomposition Method 2. We need to find factors of 12 1, 12 2, 6 3, 4 1. Find the product of first and last terms is 8 whose sum 3. Rewrite the middle term decomposed into the two numbers 3x2 + 2x + 6x + 4 = (3x2 + 2x) + (6x + 4) 4. Factor by grouping in pairs = x(3x + 2) + 2(3x + 2) = (3x + 2) (x + 2) 3x2 + 8x + 4 = (3x + 2) (x + 2) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Factoring Polynomials of the Form ax2 + bx + c

Example: Factor 4x2 + 8x – 5. 4  5 = 20 1, 20 2, 10 4, 5 We need to find factors of 20 1, 20 2, 10 4, 5 is 8 whose difference Rewrite the middle term decomposed into the two numbers 4x2 – 2x + 10x – 5 = (4x2 – 2x) + (10x – 5) Factor by grouping in pairs = 2x(2x – 1) + 5(2x – 1) = (2x – 1) (2x + 5) 4x2 + 8x – 5 = (2x –1)(2x – 5) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Example: Factor

Solve by Factoring To solve a quadratic by factoring you first get the equation equal to zero and then factor. Once factored you set each expression equal to zero and solve for the variable. Copyright © by Houghton Mifflin Company, Inc. All rights reserved.

Solving Quadratic Equations by Completing the Square

Perfect Square Trinomials Examples x2 + 6x + 9 x2 - 10x + 25 x2 + 12x + 36

Creating a Perfect Square Trinomial In the following perfect square trinomial, the constant term is missing. X2 + 14x + ____ Find the constant term by squaring half the coefficient of the linear term. (14/2)2 X2 + 14x + 49

Perfect Square Trinomials Create perfect square trinomials. x2 + 20x + ___ x2 - 4x + ___ x2 + 5x + ___ 100 4 25/4

Solving Quadratic Equations by Completing the Square Solve the following equation by completing the square: Step 1: Move quadratic term, and linear term to left side of the equation

Solving Quadratic Equations by Completing the Square Step 2: Find the term that completes the square on the left side of the equation. Add that term to both sides.

Solving Quadratic Equations by Completing the Square Step 3: Factor the perfect square trinomial on the left side of the equation. Simplify the right side of the equation.

Solving Quadratic Equations by Completing the Square Step 4: Take the square root of each side

Solving Quadratic Equations by Completing the Square Step 5: Set up the two possibilities and solve

Completing the Square-Example #2 Solve the following equation by completing the square: Step 1: Move quadratic term, and linear term to left side of the equation, the constant to the right side of the equation.

Solving Quadratic Equations by Completing the Square Try the following examples. Do your work on your paper and then check your answers.

The Quadratic Formula.

What Does The Formula Do ? The Quadratic formula allows you to find the roots of a quadratic equation (if they exist) even if the quadratic equation does not factorise. The formula states that for a quadratic equation of the form : ax2 + bx + c = 0 The roots of the quadratic equation are given by :

Example 1 Use the quadratic formula to solve the equation : x 2 + 5x + 6= 0 Solution: x 2 + 5x + 6= 0 a = 1 b = 5 c = 6 x = - 2 or x = - 3 These are the roots of the equation.

Example 2 Use the quadratic formula to solve the equation : 8x 2 + 2x - 3= 0 Solution: 8x 2 + 2x - 3= 0 a = 8 b = 2 c = -3 x = ½ or x = - ¾ These are the roots of the equation.

Example 3 Use the quadratic formula to solve the equation : Use the quadratic formula to solve the equation : 8x 2 - 22x + 15= 0 8x 2 - 22x + 15= 0 Solution: Solution: 8x 2 - 22x + 15= 0 8x 2 - 22x + 15= 0 a = 8 b = -22 c = 15 a = 8 b = -22 c = 15 x = 3/2 or x = 5/4 x = 3/2 or x = 5/4 These are the roots of the equation. These are the roots of the equation.

Example 4 Use the quadratic formula to solve for x to 2 d.p : 2x 2 +3x - 7= 0 Solution: 2x 2 + 3x – 7 = 0 a = 2 b = 3 c = - 7 x = 1.27 or x = - 2.77 These are the roots of the equation.