QUESTION concerning POLYNOMIALS....?

Question

It says this:

http://img222.imageshack.us/img222/1959/poy14re3.png

How can I solve it? I've been trying really hard, but with no luck...

Answer 1

If you have the roots, then you can find the factors by...

(x - root)

So the polynomial is (x + 5)(x + 3)(x - 2)(x - 7)

If you multiply it out...

x^4 - x^3 - 43x^2 - 23x + 210

Answer 2

Remember that, to find the zeros of a polynomial, you usually would factor it into a form like (x-a)(x-b)..., etc. Then you would find that a,b, etc. are the zeros of the polynomial.

You can use this fact to work in the other direction as well. If your degree 4 polynomial needs to have zeros at -5, -3, 2, and 7, then its factored form should be (up to a constant)
(x+5)(x+3)(x-2)(x-7).

If you want the x^4 coefficient to be 1, then this is exactly the polynomial you are looking for. As far as I'm concerned, the factored form is perfectly fine as a final answer to this question. But if you prefer, you can expand out the parentheses to get something of the form x^4 + ax^3 + ...

Answer 3

If the roots are -5, -3, 2, and 7, then you can easily find the polynomial:

f(x) = (x - (-5))(x - (-3))(x - 2)(x - 7)
=
(x+5)(x+3)(x-2)(x-7)
=
(x²+8x+15)(x²-9x+14)
=
x^4 - 9x³ + 14x² + 8x³ - 72x² + 112x + 15x² - 135x + 210
=
x^4 + x³(-9+8) + x²(14-72+15) + x(112-135) + 210
=
x^4 - x³ - 43x² - 23x + 210

Answer 4

Use the factor theorem: If r is a zero of a polynomial
then x-r is a factor.
So the answer to your question
is (x+5)(x+3)(x-2)(x-7) = x^4-x³-43x²-23x+210.

Answer 5

(x+5(x+3(x-2)(x-7=0

Answer 6

1(x+2)(x+3)(x+5)(x-7)=0
(x^2+5x+6)(x^2-2x-35)=0
x^4+3x^3-39x^2-187x-210=0