Copyright © Cengage Learning. All rights reserved. 4 Quadratic Functions Copyright © Cengage Learning. All rights reserved.

4.5 Solving Equations by Factoring Copyright © Cengage Learning. All rights reserved.

Objectives Solve a polynomial equation by factoring. Find a quadratic equation from a graph.

Solving by Factoring

Solving by Factoring Factoring can be used to solve some polynomial equations. The basis of factoring is to take a polynomial and break it into simpler pieces that are multiplied together. Standard Form Factored Form f (x) = x2 + 8x + 15 f (x) = (x + 3)(x + 5) If we multiply the factored form out and simplify, it will become the standard form.

Solving by Factoring f (x) = (x + 3)(x + 5) f (x) = x2 + 5x + 3x + 15 The main reason factoring an expression is useful is the product property of zero, which says that when you multiply any number by zero, the answer is zero.

Solving by Factoring The zero factor property follows from the product property of zero and states that if two numbers are multiplied together and the answer is zero, then one or both of those factors must be zero.

Solving by Factoring If we factor a polynomial into simpler parts, we will be able to solve the simpler equations. It is very important to note that we can use the factored form to solve an equation only if it is set equal to zero. If the quadratic equation is not equal to zero, we must first move everything to one side so that it equals zero and then factor.

Example 1 – Solving polynomial equations by factoring Solve the following by factoring. a. x2 + 7x – 40 = 20 b. 2x2 + 27 = 21x c. x3 + 8x2 + 15x = 0 Solution: x2 + 7x – 60 = 0 Set the quadratic equal to zero and factor. Steps 2 and 3 are in the margin on the right.

Example 1 – Solution x2 – 5x + 12x – 60 = 0 (x2 – 5x) + (12x – 60) = 0 cont’d x2 – 5x + 12x – 60 = 0 (x2 – 5x) + (12x – 60) = 0 x(x – 5) + 12(x – 5) = 0 (x – 5)(x + 12) = 0 x – 5 = 0 x + 12 = 0 x = 5 x = –12 Step 4. Step 5. Set the factors equal to zero and finish solving. Check both answers.

Example 1 – Solution (5)2 + 7(5) – 40 20 20 = 20 cont’d (5)2 + 7(5) – 40 20 20 = 20 (–12)2 + 7(–12) – 40 20 Both answers work.

Example 1 – Solution b. 2x2 + 27 = 21x 2x2 – 21x + 27 = 0 cont’d b. 2x2 + 27 = 21x 2x2 – 21x + 27 = 0 2x2 – 3x – 18x + 27 = 0 (2x2 – 3x) + (–18x + 27) = 0 x(2x – 3) – 9(2x – 3) = 0 (2x – 3)(x – 9) = 0 Set the quadratic equal to zero and factor. Steps 2 and 3 are in the margin on the right. Step 4. Step 5. Set the factors equal to zero and finish solving.

Example 1 – Solution 2x – 3 = 0 x – 9 = 0 x = 9 31.5 = 31.5 cont’d 2x – 3 = 0 x – 9 = 0 x = 9 31.5 = 31.5 2(9)2 + 27 21(9) 189 = 189 Check both answers. Both answers work.

Example 1 – Solution c. x3 + 8x2 + 15x = 0 x(x2 + 8x + 15) = 0 cont’d c. x3 + 8x2 + 15x = 0 x(x2 + 8x + 15) = 0 x(x2 + 3x + 5x + 15) = 0 x[(x2 + 3x) + (5x + 15)] = 0 x[x(x + 3) + 5(x + 3)] = 0 x(x + 3) + (x + 5) = 0 Take out the x in common and factor the remaining quadratic. The factored-out x will remain at the front of each step until the factoring is complete.

Example 1 – Solution x = 0 x + 3 = 0 x + 5 = 0 x = 0 x = –3 x = –5 cont’d x = 0 x + 3 = 0 x + 5 = 0 x = 0 x = –3 x = –5 (0)3 + 8(0)2 + 15(0) 0 0 = 0 (–3)3 + 8(–3)2 + 15(–3) 0 Now rewrite into three equations and solve each separately. Check all three answers in the original equation.

Example 1 – Solution (–5)3 + 8(–5)2 + 15(–5) 0 0 = 0 cont’d (–5)3 + 8(–5)2 + 15(–5) 0 0 = 0 All three answers work.

Finding an Equation from the Graph

Finding an Equation from the Graph The zero factor property can also be used to help find equations by looking at their graphs. As we solved different equations by factoring, we always set the equation equal to zero and then factor. If we set a function equal to zero and solve, we are looking for the horizontal intercepts of the function.

Finding an Equation from the Graph When we solve using factoring, we are actually finding the horizontal intercepts of the graph related to that equation. We can use the connection between the horizontal intercepts and the factors of an expression to find an equation using its graph. The horizontal intercepts of a graph also represent the zeros of the function.

Finding an Equation from the Graph In this graph, the horizontal intercepts are (–2, 0) and (–8, 0), so the zeros of the equation are and x = –2 and x = –8. Taking the zeros and working backwards through the process of solving by factoring, we will get an equation for the graph. x = –8 x = –2 x + 8 = 0 x + 2 = 0 0 = (x + 8)(x + 2) y = (x + 8)(x + 2)

Finding an Equation from the Graph This equation in factored form could be missing a constant that was factored out. To find this constant, we can use another point from the graph and solve for the missing constant. If we let a represent the missing constant and use the point (–3, 5), we can solve for a. y = a(x + 8)(x + 2)

Finding an Equation from the Graph Substituting in the –1 for a and simplifying, we get y = –1(x + 8)(x + 2) y = –x2 – 10x – 16

Finding an Equation from the Graph

Example 4 – Finding a quadratic equation from the graph Use the graph to find an equation for the quadratic. a. b. c.

Example 1(a) – Solution Step 1: Find the horizontal intercepts of the graph and write them as zeros of the equation. The horizontal intercepts for this graph are located at (–4, 0) and (3, 0), so we know that x = –4 and x = 3 are the zeros of the quadratic equation.

Example 1(a) – Solution cont’d Step 2: Undo the solving process to find the factors of the equation. Therefore, we have: x = –4 x = 3 x + 4 = 0 x – 3 = 0 This gives us the factors x + 4 and x – 3 and the equation y = (x + 4) (x – 3)

Example 1(a) – Solution cont’d Step 3: Use another point from the graph to find any constant factors. Using the point (1, –10), we get y = a(x + 4)(x – 3) –10 = a(1 + 4)(1 – 3) –10 = a(5)(–2) –10 = –10a 1 = a The constant factor is 1.

Example 1(a) – Solution cont’d Step 4: Multiply out the factors and simplify. Therefore, we have the equation y = 1(x + 4)(x – 3) y = x2 + x – 12 The y-intercept for this quadratic would be (0, –12), which agrees with the graph, so we have the correct equation.

Example 1(b) – Solution cont’d Step 1: Find the horizontal intercepts of the graph and write them as zeros of the equation. The horizontal intercepts for this graph are (2, 0) and (5, 0), so we know that x = 2 and x = 5 are the zeros of the quadratic equation.

Example 1(b) – Solution cont’d Step 2: Undo the solving process to find the factors of the equation. Therefore, we have x = 2 x = 5 x – 2 = 0 x – 5 = 0 This gives us the factors (x – 2) and (x – 5) and the equation y = (x – 2)(x – 5)

Example 1(b) – Solution cont’d Step 3: Use another point from the graph to find any constant factors. Using the point (7, 10), we get y = a(x – 2)(x – 5) 10 = a(7 – 2)(7 – 5) 10 = a(5)(2) 10 = 10a 1 = a The constant factor is 1.

Example 1(b) – Solution cont’d Step 4: Multiply out the factors and simplify. Therefore, we have the equation y = 1(x – 2)(x – 5) y = x2 – 7x + 10 The y-intercept for this quadratic would be (0, 10), which agrees with the graph, so we have the correct equation.

Example 1(c) – Solution cont’d Step 1: Find the horizontal intercepts of the graph and write them as zeros of the equation. The horizontal intercepts for this graph are (2, 0) and (5, 0), so we know that x = 2 and x = 5 are the zeros of the quadratic equation.

Example 1(c) – Solution cont’d Step 2: Undo the solving process to find the factors of the equation. Therefore, we have x = 2 x = 5 x – 2 = 0 x – 5 = 0 This gives us the factors (x – 2) and (x – 5) and the equation y = (x – 2)(x – 5)

Example 1(c) – Solution cont’d Step 3: Use another point from the graph to find any constant factors. Using the point (10, –80), we get y = a(x – 2)(x – 5) –80 = a(10 – 2)(10 – 5) –80 = 40a –2 = a The constant factor is –2.

Example 1(c) – Solution cont’d Step 4: Multiply out the factors and simplify. Therefore, we have the equation y = –2(x – 2)(x – 5) y = –2x2 + 14x – 20 The y-intercept for this quadratic would be (0, –20), which agrees with the graph, so we have the correct equation.