Here we have a quadratic trinomial of the form

a x^{2} + b x + c,

where

∣ a ∣ \neq = 1

and there are no common factors. To factor this expression, we will rewrite the middle term,

b x,

as two terms. The coefficients of these two terms will be factors of

a c

whose sum must be

b .

2 x^{2} + x - 28 \Leftrightarrow 2 x^{2} + 1 x + (- 28)

We have that

a = 2,

b = 1,

and

c = - 28 .

There are now three steps we need to follow in order to rewrite the above expression.

Find $a c .$ Since we have that $a = 2$ and $c = - 28,$ the value of $a c$ is $2 \times - 28 = - 56 .$
Find factors of $a c .$ Since $a c = - 56,$ which is negative, we need factors of $a c$ to have opposite signs — one positive and one negative — in order for the product to be negative. Since $b = 1,$ which is positive, the absolute value of the positive factor will need to be greater than the absolute value of the negative factor, so that their sum is positive.

\underline{1^{st} Factor} - 1 - 2 - 4 - 7 \underline{2^{nd} Factor} 5628148 \underline{Sum} - 1 + 56 - 2 + 28 - 4 + 14 - 7 + 8 \underline{Result} 5526101

Rewrite $b x$ as two terms. Now that we know which factors are the ones to be used, we can rewrite $b x$ as two terms.
$2 x^{2} + 1 x - 28 ⇕ 2 x^{2} - 7 x + 8 x - 28$

Finally, we will factor the last expression obtained.

2 x^{2} - 7 x + 8 x - 28

FactorOut

Factor out $x$

x (2 x - 7) + 8 x - 28

FactorOut

Factor out $4$

x (2 x - 7) + 4 (2 x - 7)

FactorOut

Factor out $(2 x - 7)$

(2 x - 7) (x + 4)

Checking Our Answer

Check your answer

✓

We can expand our answer and compare it with the given expression.

(2 x - 7) (x + 4)

Distr

Distribute $(x + 4)$

2 x (x + 4) - 7 (x + 4)

Distr

Distribute $2 x$

2 x^{2} + 8 x - 7 (x + 4)

Distr

Distribute $- 7$

2 x^{2} + 8 x - 7 x - 28

SubTerm

Subtract term

2 x^{2} + x - 28

We can see above that after expanding and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!

Exercise