Algebra 2!! help plz!!!!?

Question

Given that the polynomial has the given root, find all roots of the polynomial.

P(x) = x^4+ 4x^3+ 2x^2 -28x- 63 ; square root of 7

Answer 1

P(x) = x^4 + 4x^3 + 2x^2 - 28x - 63

sqrt(7) is a given root.

Note that we have integer coefficients; whenever we have integer coefficients and a radical root, radical roots come in conjugate pairs. sqrt(7) is a root, which means -sqrt(7) is also a root.

It also follows that (x - sqrt(7)) is a factor, and subsequently,
(x + sqrt(7)) is a factor. Multiplied together,
(x^2 - 7) is a factor.

Now, use synthetic long division to get the other two roots.
x^2 - 7 into x^4 + 4x^3 + 2x^2 - 28x - 63.

Since long division is difficult to show on here, I can tell you that the result should be x^2 + 4x + 9 with a remainder of 0. That is, if

x^4 + 4x^3 + 2x^2 - 28x - 63 = 0, then this factors as

(x^2 - 7)(x^2 + 4x + 9) = 0

We can then get the other roots using the quadratic formula.

x^2 + 4x + 9 = 0

x = [-4 +/- sqrt(4^2 - 4(9)(1))]/2
x = [-4 +/- sqrt(16 - 36)] / 2

As you can see, we're going to have imaginary roots.

x = [-4 +/- sqrt(-20)]/2
x = [-4 +/- 2sqrt(5)i]/2
x = -2 +/- sqrt(5)i

x = {sqrt(7), -sqrt(7), -2 + sqrt(5)i, -2 - sqrt(5)i}

Answer 2

If one of the roots is sqrt(7), then another root must be -sqrt(7). That means that we can factor out an (x^2-7) from the full equation. Use long division to factor this out and then you can find the other roots when the equation is quadratic.

I hope that this helps.

Answer 3

yet another root is -sqrt7.[P(-sqrt7)=0]
therefore there are 2 roots: + and - sqrt 7:
2 more roots to go:
Divide the polynomial by x^2 - 7 to get : x^2 + 4x + 9.
Roots of this quadratic are the remaining roots of the polynomial...