What polynomial has roots of -6, -4, and 1?

Question

What polynomial has roots at -6, -4, and 1? I attempted to expand the polynomial using these roots, but my calculations seem to have led to mixed results. Can someone help me clarify the correct polynomial? The expression I have is as follows: x^3 - 9x^2 - 22x + 24x^3 - x^2 - 26x - 24x^3 + x^2 - 26x + 24x^3 + 9x^2 + 14x - 24. Thank you for your assistance!

Anonymous · Accepted Answer

Write down the roots:
x = -6
or
x = -4
or
x = 1
Rewrite them as expressions equal to zero:
x + 6 = 0
or
x + 4 = 0
or
x - 1 = 0
Multiply them together to form the polynomial:
f(x) = (x + 6)(x + 4)(x - 1)
Now just expand that:
f(x) = (x² + 10x + 24)(x - 1)
f(x) = x(x² + 10x + 24) - 1(x² + 10x + 24)
f(x) = x^3 + 10x² + 24x - x² - 10x - 24
f(x) = x^3 + 9x² + 14x - 24
Answer:
D. x^3 + 9x² + 14x - 24

Anonymous · Answer

I think the point here is to figure this out without resorting to the brute force method of
mutiplying out the factors.  Knowing that roots of -6, -4, and 1 means that you will have
factors of (x + 6)( x + 4) and (x - 1).  You don't have to do ALL the multilpications to figure out
that the constant you'd get by mutiplying out the factors is going to be -24, and this allows
you to elminate two of the answers right away.
Nice thing about having a root of 1, is that's its really easy to check
x^3-x^2-26x-24 for x = 1 is 1 - 1 - 26 - 24, and that's certainly not zero, so this one doesn't
work.
The only possible answer left is the last one.

Anonymous · Answer

You could multiply out the cubic (x + 6)(x + 4)(x - 1) and see which choice fits. However, there is an easier way.....
The polynomial will have a first term of x^3 and a last (numerical) term of -24.
Only one of the given choices fits:  x^3 + 9x^2 + 14x - 24.

Anonymous

What polynomial has roots of -6, -4, and 1?

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