Problem
Name 8-2 \( \frac{\text { Reteach to Build Understanding }}{\text { Solve Quadratic Equations by Factoring }} \) 1. Match each equation with its factored form. Then match each factored form with its solution. \[ \begin{array}{lll} x^{2}+2 x-3=0 & (x-5)(x-2)=0 & \text { Solutions: }-6 \text { and }-4 \\ x^{2}+10 x+3=-21 & (x-1)(x+3)=0 & \text { Solutions: }-3 \text { and } 1 \\ x^{2}-7 x-3=-13 & (x+4)(x+6)=0 & \text { Solutions: } 2 \text { and } 5 \end{array} \] 2. Nora made an incorrect statement when using factoring to solve the equation \( x^{2}+2 x-12=3 \). Put an \( X \) next to the incorrect statement. Correct her error. The standard form of the equation is \( x^{2}+2 x-15=0 \). The factored form of the equation is \( (x-3)(x+5)=0 \). Because \( (x-3)(x+5)=0 \), you can use the Zero-Product Property to write \( (x-3)=0 \) or \( (x+5)=0 \) The solutions of the equation are -3 and 5 . The \( x \)-coordinate of the vertex of the related function is -1 . 3. Write the factored form of the equation \( x^{2}-4 x+3=15 \). Then find the solutions. Write the equation in standard form. \( \quad x^{2}-4 x \quad=0 \) Factor the quadratic equation. \( (x \quad) \quad)(x \quad)=0 \) Determine the solutions. \( (x \quad)=0 \) or \( (x \quad) \quad=0 \) \[ x=\quad x= \] The solutions of the equation \( x^{2}-4 x+3=15 \) are and enVision \( { }^{\circledast} \) Florida B.E.S.T. Algebra 1 - Teaching Resources
Solution
Step 1 ：1. \[\begin{array}{lll} x^{2}+2 x-3=0 & (x-1)(x+3)=0 & \text { Solutions: } 1 \text { and } -3 \\ x^{2}+10 x+3=-21 & (x+2)(x+4)=0 & \text { Solutions: } -2 \text { and } -4 \\ x^{2}-7 x-3=-13 & (x-5)(x-2)=0 & \text { Solutions: } 2 \text { and } 5 \end{array}\]
Step 2 ：2. X The standard form of the equation is \( x^{2}+2 x-15=0 \). Correction: The standard form of the equation is \( x^{2}+2 x-9=0 \).
Step 3 ：3. Write the equation in standard form. \( \quad x^{2}-4 x-12=0 \) Factor the quadratic equation. \( (x-6)(x+2)=0 \) Determine the solutions. \( x-6=0\text { or } x+2 =0 \) \[ x=6, x=-2 \] The solutions of the equation \( x^{2}-4 x+3=15 \) are 6 and -2
