Question

Factor each polynomial.

$x^2+4x-21$

Short answer

The factors of the given polynomial are $\left(x+3\right)\left(x-7\right)$.

Step-by-step solution

Step 1. Compare.

Compare a given trinomial to $a{x}^2+bx+c$and write the values of $a,b$and $c$.

$$\begin{gathered} a=1 \\ b=4 \\ c=-21 \end{gathered}$$

Step 2. List the factors.

List the factors of $-21$, where one factor of each pair is negative. Look for the pair of factors with a sum of 4.

Factors of$-21$	Sum of factors
$1,-21$	$-20$
$-3,7$	4

Therefore, correct factors are$-3$ and 7.

Step 3. Apply factoring pattern of ax2+bx+c.

There are two numbers having a product of $-21$and the sum of 4. The numbers are $-3$and 7.

$$\begin{gathered} x^2+4x-21=\left(x+m\right)\left(x+p\right) \\ =\left(x-3\right)\left(x+7\right) \end{gathered}$$

Therefore, the factors of$x^2+4x-21$ are $\left(x-3\right)\left(x+7\right)$.