OpenStudy (anonymous):

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

4 years ago
OpenStudy (precal):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

@precal

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (whpalmer4):

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.

4 years ago
OpenStudy (whpalmer4):

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

4 years ago
OpenStudy (anonymous):

I am just confused about the multiple choice.

4 years ago
OpenStudy (whpalmer4):

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

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

you get 24

4 years ago
OpenStudy (whpalmer4):

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?

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (anonymous):

-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!

4 years ago