Q. 361 - Intermediate Algebra

Question

In the following examples, factor each trinomial of the form $x^2+bxy+c{y}^2$

$a^2+4ab-21b^2$

Step-by-Step Solution

Answer

The factors of the polynomial $a^2+4ab-21b^2$ is $(a-3b)(a+7b)$

Step 1. Given

The trinomial is $a^2+4ab-21b^2$

To find the factors of the trinomial.

Step 2. Write binomial with first term a and second term b

The binomial contains $a$ in the first term and $b$ in the second term.

The last term of the trinomial is negative, so the factors must have opposite signs with greater number having positive sign.

$a^2+4ab-21b^2=(a+ b)(a- b)$

Step 3. Find numbers multiply to - 21 and add to 4

Find the numbers multiply to $-21$ and add to $4$.

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

Step 4. Find the factors

Use $-3,7$ as coefficients of last term

$a^2+4ab-21b^2=(a-3b)(a+7b)$

Step 5. Check the answer

The factors are $(a-3b)(a+7b)$

Multiply the factors 

$(a-3b)(a+7b)=a^2+7ab-3ab-21b^2$

                          $=a^2+4ab-21b^2$

Hence the multiplication of factors gives the trinomial.