For the polynomial function a(x)=x^4-2x^3-21x^2+22x+40 we know(x+5)is a factor.

Question

For the polynomial function  we know  is a factor. Select all the other linear factors of A(x).

A: (x + 1)

B: (x - 1)

C: (x + 2)

D: (x - 2)

E: (x + 4)

F: (x - 4)

G: (x + 8)

Answer 1

If we "know" x+5 is a factor, then x=-5 would be a zero or solution, but if you plug x=-5 into the polynomial, it doesn't = 0, so x+5 can't be a factor. Try dividing x=5 into the polynomial, it won't come out even.

But x+1 is a factor, x-1 is not. plug them into the 4th degree polynomial & see if they =0

Divide x+1 into the 4th degree and you get x^3 -3x^2 - 18x +40, with no remainder

x= 2 works, so x-2 is another factor. Divide x-2 into the cubic polynomial to get a quadratic

x^2 -x +20

which factors into (x-5)(x+4)

the four factors of the 4th degree are x-5, x+4, x+1, and x-2

Answer 2

The problem is mis-stated. The ONE given linear factor is (x - 5). NOT (x + 5).

Start by dividing the polynomial by the corrected, given factor. Use synthetic division.

Note: If (x - 5) is a linear factor, then x=5 is a zero.

And, the polynomial should be divisible by the given linear factor, with no remainder.

5. 1 -2 -21 22 40

5 15 -30 -40

-----------------------------------------

1 3 -6 -8 0.

P(x) = (x-5)(x^3 + 3x^2 - 6x - 8)

Now, factor the reduced, 3rd degree polynomial.

Possible zeros are factors are factors of 8 = +/- 1, 2, 4, 8

Try, x = -1 to see if it is a zero.

P(x) = (x-5)(x^3 + 3x^2 - 6x - 8)

P(-1) = (-6)(-1 + 3 + 6 - 8)

P(-1) = (-6)( 0 ) = 0 ==> x = -1 is a zero

Therefore, (x+1) is a linear factor.

Use synthetic division to divide

the reduced polynomial by (x+1)

-1 1 3 -6 -8

-1 -2 8

----------------------------

1 2 -8 0

P(x) = (x-5)(x+1)(x^2 + 2x - 8)

The reduced polynomial is a 2nd degree polynomial

and can easily be factored.

P(x) = (x-5)(x+1)(x+4)(x-2)

All of the linear factors of P(x) are listed above.

Answer 3

x+5 is not a factor of this expression since when you divide by (x+5) you have a remainder. For it to be a factor the remainder must be 0