Having a problem factoring a quadratic

Question

$15x^2-4x-4$, I factored it out to this: $$5x(3x-2)+2(3x+2).$$ But I don’t know what to do next since the twos in the brackets have opposite signs, or is it still possible to factor them out?

Answer 1

To split the linear term of $15x^2 - 4x - 4$, you must find two numbers with product $15 \cdot (-4) = -60$ and sum $-4$. They are $-10$ and $6$. Hence, \begin{align*} 15x^2 - 4x - 4 & = 15x^2 - 10x + 6x - 4 && \text{split the linear term}\\ & = 5x(3x - 2) + 2(3x - 2) && \text{factor by grouping}\\ & = (5x + 2)(3x - 2) && \text{extract the common factor} \end{align*} You made a sign error while extracting the factor of $2$ from $6x - 4$.

Answer 2

I think you made an arithmetic mistake. I got $$15x^2-4x-4=(5x+2)(3x-2)=5x(x-2)+2(3x-2).$$

Answer 3

$$15x^2-4x-4=15x^2-10x+6x-4=5x(3x-2)+2(3x-2)=(5x+2)(3x-2).$$

Answer 4

It has to be $-2$: $$15x^2-4x-4 = 5x(3x-2)+2(3x-2) = (5x+2)(3x-2)$$

Answer 5

Note that: $$15x^2-4x-4=15x^2-10x+6x-4=5x(3x-2)+2(3x-2)=(5x+2)(3x-2)$$

You probably missed a sign! ;)