# Factor` 4x^3-12x^2-37x-15` knowing 2x+1 is a factor. Please help!!! I really need the answer by Thursday at the most!! Thank you.

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`4x^3-12x^2-37x-15`

Since one of factors is 2x + 1, we may do synthetic division to determine the other factors.

In synthetic division, the dividend are the coefficients and constant of the given polynomial and the divisor is the root of one of the factors.

For 2x+1, the root is x = -1/2.

So,

-1/2 | 4 -12 -37 -15

-2 7 15

______________________

4 -14 -30 0

This means that,

`4x^3-12x^2-37x - 15 = (2x+1)(4x^2-14x-30)`

Next, factor out the GCF in the `4x^2-14x-30` which is 2.

`=2(2x+1)(2x^2-7x-15)`

Then, let's factor further `2x^2-7x-15` using the ac method.

To do so, multiply the coefficient of `x^2` and the constant (`2*(-15)=-30` ).

And, determine the pair factor of -30, that would have a sum equal to -7.

So the pair factor is -10 and 3.

Then, let's re-write -7x as -10x + 3x.

`2x^2-7x-15= 2x^2-10x + 3x - 15`

Group the expression into two.

`2x^2-7x-15=(2x^2-10x) + (3x-15)`

Factor out the GCF in each group.

`2x^2-7x-15=2x(x - 5) + 3(x - 5)`

And factor out the GCF too between the two terms.

`2x^2-7x-15=(x -5)(2x + 3)`

So,

`2(2x+1)(2x^2-7x-15) = 2(2x+1)(x-5)(2x+3)` **Hence, `4x^3 - 12·x^2 - 37·x - 15 = 2(2x+1)(x-5)(2x+3)` .**

Since `(2x-1)` is a factor the other factor will be a quadratic function.

Let it be `Ax^2+Bx+C`

Then we can write;

`4x^3-12x^2-37x-15 = (2x+1)(Ax^2+Bx+C)`

`4x^3-12x^2-37x-15 = 2Ax^3+2Bx^2+2Cx+Ax^2+Bx+C`

`4x^3-12x^2-37x-15 = 2Ax^3+(2B+A)x^2+(2C+B)x+C`

Now let us compare components in either side

x^3 : `4 = 2Ararr A = 2`

x^2 : `-12 = (2B+A) rarr B = (-12-2)/2 = -7`

x : `-37 = (2C+B) rarr C = (-37+7)/2 = -15`

`4x^3-12x^2-37x-15 = (2x+1)(Ax^2+Bx+C)`

`4x^3-12x^2-37x-15 = (2x+1)(2x^2-7x-15)`

since `(2x^2-7x-15)` is quadratic function we can further factor it.

`(2x^2-7x-15)`

`= 2x^2-10x+3x-15`

`= 2x(x-5)+3(x-5)`

`= (x-5)(2x+3)`

Therefore by factoring we can write;

`4x^3-12x^2-37x-15 = (2x+1)(x-5)(2x+3)`

** So **`(2x+1),((2x+3)` and `(x-5)`

**`4x^3-12x^2-37x-15` .**

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