Question8. Factor $3 x^{3}-4 x^{2}-28 x-16$ completely using long division if $(x+2)$ is a

Studdy Solution

STEP 1

What is this asking? We're asked to completely factor a cubic expression, and we're given a helpful hint that $(x+2)$ is one of the factors! Watch out! Don't forget to factor the result of the long division as much as possible.
We're looking for a completely factored form!

STEP 2

1. Set up the long division
2. Perform the long division
3. Factor the quadratic

STEP 3

Alright, let's set up our long division problem!
We're dividing $3x^3 - 4x^2 - 28x - 16$ by $(x+2)$ .
Remember, this is like dividing numbers, but with x’s!

STEP 4

First, we ask ourselves: what do we multiply $x$ by to get $3x^3$ ?
That's right, $3x^2$ !
So, we write $3x^2$ above the $x^2$ term in the dividend.

STEP 5

Now, multiply $3x^2$ by $(x+2)$ to get $3x^3 + 6x^2$ .
We subtract this from $3x^3 - 4x^2$ to get $-10x^2$ .
Don't forget to bring down the next term, $-28x$ , so we have $-10x^2 - 28x$ .

STEP 6

Next, what do we multiply $x$ by to get $-10x^2$ ?
It's $-10x$ !
Write that above the $-28x$ term.

STEP 7

Multiply $-10x$ by $(x+2)$ to get $-10x^2 - 20x$ .
Subtract this from $-10x^2 - 28x$ to get $-8x$ .
Bring down the last term, $-16$ , to get $-8x - 16$ .

STEP 8

Finally, what times $x$ gives us $-8x$ ?
It's $-8$ !
Write that above the $-16$ term.

STEP 9

Multiply $-8$ by $(x+2)$ to get $-8x - 16$ .
Subtract this from $-8x - 16$ and we get zero!
This means $(x+2)$ divides evenly into $3x^3 - 4x^2 - 28x - 16$ , with a quotient of $3x^2 - 10x - 8$ .

STEP 10

Now, we need to factor $3x^2 - 10x - 8$ .
We're looking for two numbers that multiply to $(3 \cdot -8) = -24$ and add to $-10$ .
Those numbers are $-12$ and $2$ .

STEP 11

Rewrite the quadratic as $3x^2 - 12x + 2x - 8$ .

STEP 12

Factor by grouping: $3x(x - 4) + 2(x - 4)$ .

STEP 13

This gives us $(3x+2)(x-4)$ .

STEP 14

So, the completely factored form of $3x^3 - 4x^2 - 28x - 16$ is $(x+2)(3x+2)(x-4)$ !
We did it!

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