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

Question

precal · Answer

p(x)=(x+6)(x-1)(x-4)

anonymous · Answer

x^3-9x^2-22x+24
x^3-x^2-26x-24
x^3+x^2-26x+24
x^3+9x^2+14x-24

anonymous · Answer

@precal

anonymous · Answer

Graph each function on a graphing calculator and see where the line passes through 0 at.

whpalmer4 · Answer

Actually, there are infinitely many polynomials with those roots:
$$P(x) = k(x+6)(x-1)(x-4)$$ where $k$ is a non-zero constant.  In this case, we know that $k = 1$ because the coefficient of $x^3$ in all of the answer choices is 1, and expanding the expression I gave will have a $kx^3$ term.

whpalmer4 · Answer

So, multiply out what @precal gave you, and find it on the list of answer choices.

anonymous · Answer

I am just confused about the multiple choice.

whpalmer4 · Answer

What do you get if you expand $$(x+6)(x-1)(x-4)$$?

anonymous · Answer

Use FOIL twice to multiply the binomials p(x)=(x+6)(x-1)(x-4) together to reach your answer.

anonymous · Answer

you get 24

whpalmer4 · Answer

If you do the work correctly, you will find it on the answer list...

Do you understand why we get the polynomial we do from the list of roots?

anonymous · Answer

is it x^3+x^2-26x+24?

anonymous · Answer

-6,1, and 4 are the roots. What must you do to X to make your roots equal 0 (the number you are finding)? You are adding 6, subtracting 1, and subtracting 4!

And yes, that is right!