How do you simplify #[(x^3 - 1)/(x + 4)][(2x - 7)/(x^2 + 3x + 1)]#?

Question

Avery Alexander · Accepted Answer

#(2x^4 - 7x^3 - 2x + 7)/(x^3 + 7x^2 + 13x + 4)#

We multiply the numerators with numerators and denominators with denominators.

First, let's look at the numerators.

We multiply the numerators like this: #(x^3-1)(2x-7)# and now we need to simplify using the rainbow, FOIL, box method, or whatever way you want to do it.

#x^3 * 2x = 2x^4#

#x^3 * -7 = -7x^3#

#-1 * 2x = -2x#

#-1 * -7 = 7#

So if we put them all together, we will get #2x^4 - 7x^3 - 2x + 7#.

Now multiply the denominators: #(x + 4)(x^2 + 3x + 1)#

#x * x^2 = x^3#

#x * 3x = 3x^2#

#x * 1 = x#

#4 * x^2 = 4x^2#

#4 * 3x = 12x#

#4 * 1 = 4#

Again, let's put them all together, and we get #x^3 + 3x^2 + x + 4x^2 + 12x + 4#

We still have to combine the "like terms": #x^3 + 7x^2 + 13x + 4#

So our final answer is: #(2x^4 - 7x^3 - 2x + 7)/(x^3 + 7x^2 + 13x + 4)#

Avery Alexander · Answer

undefined To simplify the expression [(x^3 - 1)/(x + 4)][(2x - 7)/(x^2 + 3x + 1)], we can first factor the numerator and denominator of each fraction.

The numerator of the first fraction, x^3 - 1, can be factored as (x - 1)(x^2 + x + 1).

The denominator of the first fraction, x + 4, cannot be factored further.

The numerator of the second fraction, 2x - 7, cannot be factored further.

The denominator of the second fraction, x^2 + 3x + 1, cannot be factored further.

Now, we can cancel out any common factors between the numerators and denominators.

(x - 1) cancels out with (x - 1) in the first fraction.

No other common factors can be canceled out.

After canceling out the common factors, the simplified expression becomes:

[(x^2 + x + 1)/(x + 4)][(2x - 7)/(x^2 + 3x + 1)]