I need help factoring this. x^5 - 15x^4 + 54x^3?

Question

I need some help factoring the question above. I am not sure how to do this. Do I break it down to the GCF first.

Answer 1

This one actually isn't too bad.

First of all, we see that every term has a number of x's in it. In fact, each has at least 3, so we can pull out an x^3.

x^5 - 15x^4 + 54x^3

x^3 (x^2 - 15x + 54)

Now, we need two numbers that add to -15 and multiply to 54, and those numbers are -6 and -9. So...

x^3 (x - 6)(x - 9)

We cannot factor this any further, so we have our final answer:

x³ (x - 6)(x - 9)

Answer 2

I would factor out x^3 first, the GCF in terms of degree of x:

x^5 - 15x^4 + 54x^3 =
x^3 (x^2 - 15x + 54)

Then, it's a simple matter of factoring the remaining quadratic equation. This one can be done in your head: you need two negative numbers whose sum is -15 and whose product is 54. 54 = 6*9, and the sum of (-6)+(-9) is -15, so that works:

x^3 (x-6) (x-9)

Answer 3

Start by factoring out x^3. You end up with x^3(x^2-15x+54). YOu can factor x^2-15x+54 to (x-6)(x-9).

Therefore:

X^3(x-6)(x-9)

Answer 4

x^5 - 15x^4 + 54x^3
x^3(x^2 - 15x + 54)
x^3(x-6)(x-9)
x = 0, 6, 9

Answer 5

x^5-15x^4+54x^3=x^3(x^2-15x+54)
Now, to factor x^2-15x+54
it is of the form ax^2+bx+c. x=[-b+or-(b^2-4ac)]/2a
=[(-15)+(225-216]/2 =-18/2=-9 or
[(-15)-(225-216)]/2=-12/2=-6
Final answer is x^3(x-9)(x-6)

Answer 6

Obviously, you can take out a factor of x^3, and the resulting quadratic is easily factored to give (x-6)(x-9)

Answer 7

x³ (x² - 15x + 54)
x³ (x - 9) (x - 6)
x = 0, 9, 6

Please rate my best

Answer 8

x^3(x-6)(x-9)