# Is it possible to factor y=3x^2+11x-4 ? If so, what are the factors?

Dec 9, 2015

$y = \left(x + 4\right) \left(3 x - 1\right)$

#### Explanation:

Use the AC method.

$y = 3 {x}^{2} + 11 x - 4$ is a quadratic equation in the form $a x + b x + c$, where $a = 3 , b = 11 , \mathmr{and} c = - 4$.

Multiply $a$ times $c$.

$3 \times - 4 = - 12$

Find two numbers that when added equal $11$ and when multiplied equal $- 12$. The numbers $12$ and $- 1$ meet the criteria.

Rewrite the equation substituting $12 x$ and $- x$ for $11 x$.

$3 {x}^{2} + 12 x - x - 4$

Group the terms into two groups of two terms.

$\left(3 {x}^{2} + 12 x\right) - \left(x - 4\right)$

Factor out $3 x$ from the first group and $- 1$ from the second group.

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

Factor out $\left(x + 4\right)$.

$\left(x + 4\right) \left(3 x - 1\right)$

$y = \left(x + 4\right) \left(3 x - 1\right)$

Dec 9, 2015

$\textcolor{b r o w n}{\left(3 x - 1\right) \left(x + 4\right)}$

#### Explanation:

3 is prime so we can only have 1 and 3 as factors giving:

(x+?)(3x+?)

There are two sets of factors of 4 and they are: { 2,2} and {1,4}

Lets try the {1,4} combination and see what we get!

By the way, the constant of 4 in your question is negative so the $\textcolor{w h i t e}{\ldots .}${1 ,4} have to be opposite in their sign.

We see that $3 \times 4 = 12$ which is close to the $11$ in $11 x$ so lets try that configuration:

$\left(x - 4\right) \left(3 x + 1\right) = 3 {x}^{2} - 12 x + x - 4$ Not quite correct as we have $- 11 x$.

Ok! Lets try something else. The 11 is the correct magnitude but the wrong sign. Lets try reversing the signs:

$\left(x + 4\right) \left(3 x - 1\right) = 3 {x}^{2} + 12 x - x - 4$ Now we have it!

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So $\textcolor{w h i t e}{. .} 3 {x}^{2} + 11 x - 4 = \textcolor{b r o w n}{\left(x + 4\right) \left(3 x - 1\right)}$

Or if you like you can change the order of the brackets to.

$\textcolor{b r o w n}{\left(3 x - 1\right) \left(x + 4\right)}$

$\textcolor{b l u e}{\text{They call this Commutative}}$

in that they can commute/travel without changing the intrinsic value