Question

(X^2+x-6)(2x^2+4x) factor completely

Answer 1

Answer:

$2x\left(x-2\right)\left(x+3\right)\left(x+2\right)$

Step-by-step explanation:

Lets go ahead and take a step by step approach to solving your problem.

First we must start by factoring $x^2+x-6$.

To do this, we must break the expression into groups! Although I will also provide a definition to help you understand! For $ax^2+bx+c$ we need to find u, v > $u\cdot \:v=a\cdot \:c$ and $u+v=b$ which we will group into $\left(ax^2+ux\right)+\left(vx+c\right)$.

The values we have are...$a=1,\:b=1,\:c=-6$ and $u\cdot v=-6,\:u+v=1$.

Now we need the factors of 6 which are 1, 2, 3 and 6. We also need the negative factors which you get simply by multiplying the positives by -1 or just reversing them.

Now we need to check for every two factors if u + v = 1.

$\mathrm{Check}\:u=1,\:v=-6: \:u\cdot v=-6,\:u+v=-5 \Rightarrow \mathrm{False}$

$\mathrm{Check}\:u=2,\:v=-3: \:u\cdot v=-6,\:u+v=-1 \Rightarrow \mathrm{False}$

$\mathrm{Check}\:u=3,\:v=-2: \:u\cdot v=-6,\:u+v=1 \Rightarrow \mathrm{True}$

$\mathrm{Check}\:u=6,\:v=-1: \:u\cdot v=-6,\:u+v=5 \Rightarrow \mathrm{False}$

Therefore $u=3,\:v=-2$.

Now we want to $\mathrm{Group\:into\:}\left(ax^2+ux\right)+\left(vx+c\right)$. Which is $\left(x^2-2x\right)+\left(3x-6\right)$.

Now we must factor x from $x^2-2x$.

Lets apply the exponent rule of $a^{b+c}=a^ba^c$ which means $x^2=xx$.

Therefore we now have xx - 2x. Lets factor out the common term of x to get x(x - 2).

Now factor 3 out of $3x-6$.

Rewrite 6 as 3 * 2. $3x-3\cdot \:2$.

Now factor out the common term of 3 > $3\left(x-2\right)$.

$x\left(x-2\right)+3\left(x-2\right)$ > Factor out the common term of x - 2 > $\left(x-2\right)\left(x+3\right)$.

Now lets factor $2x^2+4x$.

Apply the previous exponent rule of $a^{b+c}=a^ba^c$ > $x^2=xx$ > $2xx+4x$.

Now rewrite 4 as 2 * 2 > $2xx+2\cdot \:2x$.

Now factor out the common term of 2x to get 2x(x + 2).

Combine it all and we get $2x\left(x-2\right)\left(x+3\right)\left(x+2\right)$.

Hope this helps!