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 .

3 x^{2} + 16 x + 5

We have that

a = 3,

b = 16,

and

c = 5 .

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

Find $a c .$ Since we have that $a = 3$ and $c = 5,$ the value of $a c$ is $3 \times 5 = 15 .$
Find factors of $a c .$ Since $a c = 15,$ which is positive, we need factors of $a c$ to have the same sign — both positive or both negative — in order for the product to be positive. Since $b = 16,$ which is also positive, those factors will need to be positive so that their sum is positive.

\underline{1^{st} Factor} 31 \underline{2^{nd} Factor} 515 \underline{Sum} 3 + 5 1 + 15 \underline{Result} 816

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.
$3 x^{2} + 16 x + 5 \Leftrightarrow 3 x^{2} + 1 x + 15 x + 5$

Finally, we will factor the last expression obtained.

3 x^{2} + 1 x + 15 x + 5

FactorOut

Factor out $x$

x (3 x + 1) + 15 x + 5

FactorOut

Factor out $5$

x (3 x + 1) + 5 (3 x + 1)

FactorOut

Factor out $(3 x + 1)$

(3 x + 1) (x + 5)

Checking Our Answer

Check your answer

✓

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

(3 x + 1) (x + 5)

Distr

Distribute $(x + 5)$

3 x (x + 5) + (x + 5)

Distr

Distribute $3 x$

3 x^{2} + 15 x + (x + 5)

RemovePar

Remove parentheses

3 x^{2} + 15 x + x + 5

AddTerms

Add terms

3 x^{2} + 16 x + 5

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!

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