Question

Use the Remainder Theorem to completely factor p(x) = x ^3 + 7x ^2 + 11x + 5.

Answer 1

In order to use the remainder theorem, you need to have some idea what to divide by. The rational root theorem tells you rational roots will be from the list derived from the factors of the constant term, {±1, ±5}. When we compare coefficients of odd power terms to those of even power terms, we find their sums are equal, which means -1 is a root and (x +1) is a factor.

Dividing that from the cubic, we get a quotient of x² +6x +5 (and a remainder of zero). We recognize that 6 is the sum of the factors 1 and 5 of the constant term 5, so the factorization is

... = (x +1)(x +1)(x +5)

... = (x +1)²(x +5)

_____

The product of factors (x +a)(x +b) will be x² + (a+b)x + ab. That is, the factorization can be found by looking for factors of the constant term (ab) that add to give the coefficient of the linear term (a+b). The numbers found can be put directly into the binomial factors to make (x+a)(x+b).

When we have 1·5 = 5 and 1+5 = 6, we know the factorization of x²+6x+5 is (x+1)(x+5).

Answer 2

Answer: (x+1)(x+1)(x+5)

Explanation:

let f(x) = x^3 + 7x^2 + 11x +5

let x= -1

f(-1)= (-1)^3 + 7(-1)^2 + 11(-1) + 5

= -1 +7-11+5

=0

Therefore x + 1 is a factor

Then we proceed to divide

x^3 + 7x^2 + 11x +5 by x+1

_x^2+6x+5____

x+1√x^3 + 7x^2 + 11x + 5

-(x^3 + x^2)

____________________

6x^2 + 11x

- (6x^2 + 6x)

_______________________

5x +5

- (5x + 5)

_______________________

0

so after division,

x^2+6x+5 is also another factor, but we can further break it down into 2 factors.

let f(x) = x^2 + 6x + 5

let x= -1

f(-1)= (-1)^2 + 6(-1) + 5

= 1 - 6 + 5

= 0

Therefore (x+1) is also another factor,

we can use x+ 1 to divide

x^2 + 6x + 5 to get the next factor

_____x+5_____

x+1√x^2+6x +5

- (x^2+6x)

_____________

5x + 5

- (5x+5)

______________

0

Therefore x + 5 is another factors

(x+1)(x+1)(x+5) are the factors of the polynomial.