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