To find the remaining factors we will use the and then, we will factor the result completely.
Synthetic division
To divide the given using , all the terms of the dividend must be present. Since there are no missing
terms, we do not need to rewrite the polynomial.
x^3 + x^2 - 16x - 16Remember that the general form of the divisor in synthetic division is x-a. Therefore, we need to rewrite ours a little bit.
x+1 ⇔ x-(-1)
Now we are ready to divide!
rl IR-0.15cm r -1 & |rr 1& 1 & - 16 &- 16
Bring down the first coefficient
rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 & c 1 & & &
Multiply the coefficient by the divisor
rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 &-1&& & c 1 & & &
rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 &-1&& & c 1 & 0 & &
â–¼
Repeat the process for all the coefficients
Multiply the coefficient by the divisor
rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 &-1& 0& & c 1 & 0 & &
rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 &-1& 0& & c 1 & 0 & -16 &
Multiply the coefficient by the divisor
rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 &-1& 0& 16 & c 1 & 0 & -16 &
rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 &-1& 0& 16 & c 1 & 0 & -16 & 0
The quotient is a polynomial of degree 2 and the remainder is zero. We know that our remainder is correct because we were told that x+1 is a factor of the polynomial. Let's rewrite the given polynomial as the product of two factors.
( x^3+x^2-16x-16 ) = (x+1) ( x^2-16)
Factoring
Finally, let's factor the quadratic polynomial. To do so, we will use .
(x+1)( x^2 - 16 )
(x+1) ( x^2 - 4^2 )
(x+1)(x+4)(x-4)
Now, we can see, that the remaining factors are x +4 and x - 4.
