# How do you factor  9x^2+24xy+16y^2?

Mar 17, 2018

$9 {x}^{2} + 24 x y + 16 {y}^{2} = {\left(3 x + 4 y\right)}^{2}$

#### Explanation:

Note that:

$9 {x}^{2} = {\left(3 x\right)}^{2}$

$24 x y = 2 \left(3 x\right) \left(4 y\right)$

$16 {y}^{2} = {\left(4 y\right)}^{2}$

So:

$9 {x}^{2} + 24 x y + 16 {y}^{2} = {\left(3 x\right)}^{2} + 2 \left(3 x\right) \left(4 y\right) + {\left(4 y\right)}^{2}$

is in the form:

${A}^{2} + 2 A B + {B}^{2} = {\left(A + B\right)}^{2}$

So putting $A = 3 x$ and $B = 4 y$ we have:

$9 {x}^{2} + 24 x y + 16 {y}^{2} = {\left(3 x + 4 y\right)}^{2}$

Mar 17, 2018

$\left(3 x + 4 y\right) \left(3 x + 4 y\right)$

#### Explanation:

We are given:

$9 {x}^{2} + 24 x y + 16 {y}^{2}$

We want to obtain an expression of the form:

$\left(a x + b y\right) \left(c x + \mathrm{dy}\right)$

where $a , b , c , d$ are integers (not necessarily unique from each other).

Expanding this form we get:

$a c {x}^{2} + a \mathrm{dx} y + b c x y + b {\mathrm{dy}}^{2}$

$a c {x}^{2} + \left(a b + c d\right) x y + b {\mathrm{dy}}^{2}$

From the expression we are given we must satisfy the following equations:

$a c = 9$
$b d = 16$
$a b + c d = 24$

For $\left(a , c\right)$ we can have (eliminating any repeats):
$\left(1 , 9\right) , \left(3 , 3\right) , \cancel{\left(9 , 1\right)}$

For $\left(b , d\right)$ we can have (eliminating any repeats):
$\left(1 , 16\right) , \left(2 , 8\right) , \left(4 , 4\right) , \cancel{\left(8 , 2\right)} , \cancel{\left(16 , 1\right)}$

With these options, we now need to find the ones that when combined will give us the $a b + c d = 24$:

$\left(1 , 9\right) \left(1 , 16\right) \to 1 \times 1 + 9 \times 16 = 145$
$\left(1 , 9\right) \left(2 , 8\right) \to 1 \times 2 + 9 \times 8 = 74$
$\left(1 , 9\right) \left(4 , 4\right) \to 1 \times 4 + 9 \times 4 = 40$
$\left(3 , 3\right) \left(1 , 16\right) \to 3 \times 1 + 3 \times 16 = 51$
$\left(3 , 3\right) \left(2 , 8\right) \to 3 \times 2 + 3 \times 8 = 30$
$\textcolor{b l u e}{\left(3 , 3\right) \left(4 , 4\right) \to 3 \times 4 + 3 \times 4 = 24}$

Hence, we must choose $\left(a , c\right) = \left(3 , 3\right)$ and $\left(b , d\right) = \left(4 , 4\right)$

So the factored expression $\left(a x + b y\right) \left(c x + \mathrm{dy}\right)$ is:

$9 {x}^{2} + 24 x y + 16 {y}^{2} = \textcolor{g r e e n}{\left(3 x + 4 y\right) \left(3 x + 4 y\right)}$