Question

What are the zeros of the polynomial function? f(x)=x^4+2x^3-16x^2-2x+15

Answer 1

(x + 1)(x - 1)(x + 5)(x - 3) is the fully factored form of the polynomial.

The zeros are (-1, 0), (1, 0), (-5, 0), and (3, 0).

x^4 + 2x^3 - 16x^2 - 2x + 15

We can use the rational roots theorem to find some of the possible roots, and after finding just one root, we can simplify this polynomial.

List factors of 15:

1, 3, 5, 15.

List factors of 1:

1.

Our possible rational factors are:

+/- 1, +/- 3, +/- 5, +/- 15.

To find factors, we can use the remainder theorem.

Replace all x values with 1.

1^4 + 2(1)^3 - 16(1)^2 - 2(1) + 15 = 0

Because the answer is zero, it means that 1 is a root.

We can divide this polynomial by x - 1 to find a simplified form.

After dividing, our quotient is: x^3 + 3x^2 - 13x - 15

We can continue finding factors by using the rational roots theorem. Once we have only three terms, we can try to factor using the AC method.

Our next possible root is -1.

(-1)^3 + 3(-1)^2 - 13(-1) - 15 = 0

We know that -1 is also a root, and so we can divide the polynomial by x + 1.

After diving we're left with x^2 + 2x - 15.

Now, we can try to factor using the AC method.

List factors of -15.

1 * -15

-1 * 15

3 * -5

-3 * 5 (these digits satisfy the criteria.)

Split the middle term.

x^2 - 3x + 5x - 15

Factor binomials.

x(x - 3) + 5(x - 3)

Rearrange binomials.

(x + 5)(x - 3)

Add in the two factors we already factored out.

(x - 1)(x + 1)(x + 5)(x - 3)

Answer 2

$x^4+2x^3-16x^2-2x+15 =0\\\\x^4+x^3+x^3+x^2-17x^2-17x+15x+15=0\\\\x^3(x+1)+x^2(x+1)-17x(x+1)+15(x+1)=0\\\\(x+1)(x^3+x^2-17x+15)=0\\\\(x+1)(x^3-x^2+2x^2-2x-15x+15)=0\\\\(x+1)(x^2(x-1)+2x(x-1)-15(x-1))=0\\\\(x+1)(x-1)(x^2+2x-15)=0\\\\(x+1)(x-1)(x^2-3x+5x-15)=0\\\\(x+1)(x-1)(x(x-3)+5(x-3))=0\\\\(x+1)(x-1)(x+5)(x-3)=0\\\\x=-1 \vee x=1 \vee x=-5 \vee x=3$