- d.First, factor the polynomial:
9 –7x– 2x^2 = 9 + 2x– 9x– 2x^2 = 1(9 + 2x)
- x(9 + 2x) = (9 + 2x)(1 – x)
The factors are 9 + 2xand 1 – x. Set each fac-
tor equal to zero and solve for x.The zeros of
the polynomial are –^92 and 1.
- b.Begin by factoring the polynomial:
2 x^3 + 6x^2 + 4x= 2x(x^2 + 3x+ 2) = 2x(x^2 + x+
2 x+ 2)
2 x(x(x+ 1) + 2(x+ 1) = 2x(x+ 2)(x+ 1)
There are three factors: 2x,x+ 2, and x+ 1.
Set each factor equal to zero and solve for xto
conclude that the zeros of the polynomial are
–2, –1, and 0.
- c.First, factor the polynomial:
–4x^5 + 24x^4 – 20x^3 = –4x^3 (x^2 – 6x+ 5) =
–4x^3 (x^2 – x– 5x+ 5)
= –4x^3 (x(x– 1) – 5(x– 1)) = –4x^3 (x– 5)(x– 1)
The three factors are –4x^3 ,x– 5, and x– 1. Now,
set each factor equal to zero and solve for xto
find that zeros of the polynomial: 0, 1, and 5.
446 a.First, factor the polynomial:
2 x^2 (x^2 – 4) – x(x^2 – 4) + (4 – x^2 ) = 2x^2 (x^2 – 4)
- x(x^2 – 4) – (x^2 – 4) = (x^2 – 4)[2x^2 – x– 1]
= (x^2 – 4)[2x^2 – 2x+ x– 1] = (x^2 – 4)[2x(x– 1)
+ (x– 1] = (x^2 – 4) (2x+1)(x– 1)
= (x– 2)(x+ 2)(2x+ 1)(x– 1)
Now, set each of the four factors equal to zero
and solve for x.The zeros of the polynomial
are 1, 2, –2, and –^12 .
- d.Begin by factoring the polynomial:
2 x^2 (16 + x^4 ) + 3x(16 + x^4 ) + (16 + x^4 ) =
(16 + x^4 )[2x^2 + 3x+ 1]
= (16 + x^4 )[2x^2 + 2x+ x+ 1] = (16 + x^4 )[2x(x+
1) + (x+ 1)] = (16 + x^4 )(2x+ 1)(x+ 1)
The three factors are 16 + x^4 ,2x+ 1, and x+ 1.
Now, set each factor equal to zero. Solve for x
to find that zeros of the polynomial: –1 and –^12 .
- b.First, factor the polynomial:
18(x^2 + 6x+ 8) – 2x^2 (x^2 + 6x+ 8) = (x^2 +
6 x+ 8)[18 – 2x^2 ] = (x^2 + 6x+ 8)[2(9 – x^2 )]
= (x^2 + 4x+ 2x+ 8)[2(3^2 – x^2 )] = (x(x+ 4) +
2(x+ 4))[2(3 – x)(3 + x)]
= 2(x+ 2)(x+ 4)(3 – x)(3 + x)
Set each of the four factors equal to zero and
solve for x.The zeros of the polynomial are
–4, –2, –3, and 3.
Set 29 (Page 75)
- b.The strategy is to determine the x-values
that make the expression on the left side equal
to zero. Doing so requires that we first factor
the polynomial:
x^2 – 36 = x^2 – 6^2 = (x– 6)(x+ 6)
Next, set each factor equal to zero and solve
for xto conclude that the zeros of the polyno-
mial are –6 and 6. Now, we assess the sign of
the expression on the left side on each subin-
terval formed using these values. To this end,
we form a number line, choose a real number
in each of the subintervals, and record the sign
of the expression above each:
–6 6
+ – +
ANSWERS & EXPLANATIONS–
