Q 269

In the following exercises, factor completely.

$9 x^{2} - 6 x y + y^{2} - 49$

Verified

Answer

The expression is factored as $9 x^{2} - 6 x y + y^{2} - 49 = (3 x - y - 7) (3 x - y + 7)$

1Step 1. Given Information

Consider the expression $9 x^{2} - 6 x y + y^{2} - 49$ .

The objective is to factor the expression completely.

2Step 2. Factor using perfect square trinomial

The trinomial $a^{2} - 2 a b + b^{2}$ is a perfect square trinomial and it is factored as ${(a - b)}^{2}$ .

In the given expression, the first three terms form a perfect square trinomial and so it can be factored as

$\begin{array}{rcl} 9 x^{2} - 6 x y + y^{2} - 49 & = & {(3 x)}^{2} - 2 \cdot 2 x \cdot y + {(y)}^{2} - 49 \\ = & {(3 x - y)}^{2} - 49 \end{array}$

3Step 3. Factor using the difference of square

The difference of squares can be factored as $a^{2} - b^{2} = (a - b) (a + b)$

Now the first term is a perfect square and the second term $49$ is the square of $7$ . So it can be further factored as

$\begin{array}{rcl} {(3 x - y)}^{2} - 49 & = & {(3 x - y)}^{2} - 7^{2} \\ = & (3 x - y - 7) (3 x - y + 7) \end{array}$

4Step 4. Check the factors

Multiply the factors to check the solution

\begin{array}{rcl} (3 x - y - 7) (3 x - y + 7) & = & 9 x^{2} - 3 x y + 21 x - 3 x y + y^{2} - 7 y - 21 x + 7 y - 49 \\ = & 9 x^{2} - 6 x y + y^{2} - 49 \end{array}

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