Q 263

Question

In the following exercises, factor completely.

$36 x^{2} y + 15 x y - 6 y$

Step-by-Step Solution

Verified

Answer

The expression is factored as $36 x^{2} y + 15 x y - 6 y = 3 y (4 x - 1) (3 x + 2)$

1Step 1. Given Information

Consider the expression $36 x^{2} y + 15 x y - 6 y$ .

We need to factor the expression completely.

2Step 2. Factor the GCF

The GCF of the three terms in the expression is $3 y$ . So factor the GCF from the expression we get

\begin{array}{rcl} 36 x^{2} y + 15 x y - 6 y & = & 3 y \cdot 12 x^{2} + 3 y \cdot 5 x - 3 y \cdot 2 \\ = & 3 y (12 x^{2} + 5 x - 2) \end{array}

3Step 3. Split the middle term to factor the trinomial

The middle term of the trinomial $5 x$ can be splitted as $8 x - 3 x$ and then it can be factored completely as

$\begin{array}{rcl} 12 x^{2} + 5 x - 2 & = & 12 x^{2} + 8 x - 3 x - 2 \\ = & 4 x (3 x + 2) - (3 x + 2) \\ = & (3 x + 2) (4 x - 1) \end{array}$

Thus, $36 x^{2} y + 15 x y - 6 y = 3 y (4 x - 1) (3 x + 2)$

4Step 4. Check the solution

Multiply the factors to check the solution

$\begin{array}{rcl} 3 y (4 x - 1) (3 x + 2) & = & 3 y (12 x^{2} + 8 x - 3 x - 2) \\ = & 3 y (12 x^{2} + 5 x - 2) \\ = & 36 x^{2} y + 15 x y - 6 y \end{array}$

And we get the given expression. So the expression is correctly factored.

Q. 262

Q 264

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