Exercise 5 Page 355

To find the remaining factors we will use the synthetic division and then, we will factor the result completely.

Synthetic division

To divide the given polynomials using synthetic division, 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 & & &

Add down

rl IR-0.15cm r -1 & |rr 1 & 1 & - 16 &- 16 &-1&& & c 1 & 0 & &

▼

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 & &

Add down

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 &

Add down

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 difference of squares rule.

(x+1)( x^2 - 16 )

WritePow

Write as a power

(x+1) ( x^2 - 4^2 )

FacDiffSquares

a^2-b^2=(a+b)(a-b)

(x+1)(x+4)(x-4)

Now, we can see, that the remaining factors are x +4 and x - 4.