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Zeros of Polynomials
In Exercises 16, use direct substitution
to show that the given value is a
zero of the given polynomial.
1. p(x) = x^{3} − 3x^{2} − 13x + 15, x = −3
2. p(x) = x^{3} − 2x^{2} − 13x − 10, x = −2
3. p(x) = x^{4} − x^{3} − 12x^{2}, x = 4
4. p(x) = x^{4} − 2x^{3} − 3x^{2}, x = −1
5. p(x) = x^{4} + x^{2} − 20, x = −2
6. p(x) = x^{4} + x^{3} − 19x^{2} + 11x + 30,
x = −1
In Exercises 728, identify all of the
zeros of the given polynomial without
the aid of a calculator . Use an algebraic
technique and show all work (factor
when necessary) needed to obtain the
zeros.
7. p(x) = (x − 2)(x + 4)(x − 5)
8. p(x) = (x − 1)(x − 3)(x + 8)
9. p(x) = −2(x − 3)(x + 4)(x − 2)
10. p(x) = −3(x + 1)(x − 1)(x − 8)
11. p(x) = x(x − 3)(2x + 1)
12. p(x) = −3x(x + 5)(3x − 2)
13. p(x) = −2(x + 3)(3x − 5)(2x + 1)
14. p(x) = 3(x − 2)(2x + 5)(3x − 4)
15. p(x) = 3x^{3} + 5x^{2} − 12x − 20
16. p(x) = 3x^{3} + x^{2} − 12x − 4
17. p(x) = 2x^{3} + 5x^{2} − 2x − 5
18. p(x) = 2x^{3} − 5x^{2} − 18x + 45
19. p(x) = x^{4} + 4x^{3} − 9x^{2} − 36x
20. p(x) = −x^{4} + 4x^{3} + x^{2} − 4x
21. p(x) = −2x^{4} − 10x^{3} + 8x^{2} + 40x
22. p(x) = 3x^{4} + 6x^{3} − 75x^{2} − 150x
23. p(x) = 2x^{3} − 7x^{2} − 15x
24. p(x) = 2x^{3} − x^{2} − 10x
25. p(x) = −6x^{3} + 4x^{2} + 16x
26. p(x) = 9x^{3} + 3x^{2} − 30x
27. p(x) = −2x^{7} − 10x^{6} + 8x^{5} + 40x^{4}
28. p(x) = 6x^{5} − 21x^{4} − 45x^{3}
In Exercises 2934, the graph of a polynomial
is given. Perform each of the following
tasks.
i. Copy the image onto your homework
paper. Label and scale your axes,
then label each x intercept with its
coordinates.
ii. Identify the zeros of the polynomial.
29.
30.
31.
32.
33.
34.
For each of the polynomials in Exercises 35
46, perform each of the following tasks.
i. Factor the polynomial to obtain the
zeros. Show your work.
ii. Set up a coordinate system on graph
paper. Label and scale the horizontal
axis. Use the zeros and endbehavior
to help sketch the graph of the polynomial
without the use of a calculator .
iii. Verify your result with a graphing calculator.
35. p(x) = 5x^{3} + x^{2} − 45x − 9
36. p(x) = 4x^{3} + 3x^{2} − 64x − 48
37. p(x) = 4x3 − 12x^{2} − 9x + 27
38. p(x) = x^{3} + x^{2} − 16x − 16
39. p(x) = x^{4} + 2x^{3} − 25x^{2} − 50x
40. p(x) = −x^{4} − 5x^{3} + 4x^{2} + 20x
41. p(x) = −3x^{4} − 9x^{3} + 3x^{2} + 9x
42. p(x) = 4x^{4} − 29x^{2} + 25
43. p(x) = −x^{3} − x^{2} + 20x
44. p(x) = 2x^{3} − 7x^{2} − 30x
45. p(x) = 2x^{3} + 3x^{2} − 35x
46. p(x) = −2x^{3} − 11x^{2} + 21x
6.2 Solutions
1. p(−3) = (−3)^{3} − 3(−3)^{2} − 13(−3) + 15 = −27 − 27 + 39 + 15 = 0
3. p(4) = 4^{4} − 4^{3} − 12(4)^{2} = 256 − 64 − 192 = 0
5. p(−2) = (−2)^{4} + (−2)^{2} − 20 = 16 + 4 − 20 = 0
7. Set p(x) = 0 in p(x) = (x − 2)(x + 4)(x − 5),
0 = (x − 2)(x + 4)(x − 5),
then use the zero product property to write
x − 2 = 0 or x + 4 = 0 or x − 5 = 0.
Solving , the zeros are x = 2, −4, and 5.
9. Set p(x) = 0 in p(x) = −2(x − 3)(x + 4)(x − 2),
0 = −2(x − 3)(x + 4)(x − 2),
then use the zero product property to write
x − 3 = 0 or x + 4 = 0 or x − 2 = 0.
Solving, the zeros are x = 3, −4, and 2.
11. Set p(x) = 0 in p(x) = x(x − 3)(2x + 1),
0 = x(x − 3)(2x + 1),
then use the zero product property to write
x = 0 or x − 3 = 0 or 2x + 1 = 0.
Solving, the zeros are x = 0, 3, and −1/2.
13. Set p(x) = 0 in p(x) = −2(x + 3)(3x − 5)(2x + 1),
0 = −2(x + 3)(3x − 5)(2x + 1),
then use the zero product property to write
x + 3 = 0 or 3x − 5 = 0 or 2x + 1 = 0.
Solving, the zeros are x = −3, 5/3, and −1/2.
15. Factor p(x) = 3x^{3} + 5x^{2} − 12x − 20 by grouping, then
finish the factoring using
the difference of squares pattern.
p(x) = 3x^{3} + 5x^{2} − 12x − 20
p(x) = x^{2}(3x + 5) − 4(3x + 5)
p(x) = (x^{2} − 4)(3x + 5)
p(x) = (x + 2)(x − 2)(3x + 5)
To find the zeros, set p(x) = 0,
0 = (x + 2)(x − 2)(3x + 5),
then use the zero product property to write
x + 2 = 0 or x − 2 = 0 or 3x + 5 = 0.
Solving, the zeros are x = −2, 2, or −5/3.
17. Factor p(x) = 2x^{3} +5x^{2} −2x−5 by grouping, then
finish the factoring using the
difference of squares pattern .
p(x) = 2x^{3} + 5x^{2} − 2x − 5
p(x) = x^{2}(2x + 5) − 1(2x + 5)
p(x) = (x^{2} − 1)(2x + 5)
p(x) = (x + 1)(x − 1)(2x + 5)
To find the zeros, set p(x) = 0,
0 = (x + 1)(x − 1)(2x + 5),
then use the zero product property to write
x + 1 = 0 or x − 1 = 0 or 2x + 5 = 0.
Solving, the zeros are x = −1, 1, or −5/2.
19. Start with p(x) = x^{4} + 4x^{3} − 9x^{2} − 36x. Factor out
the gcf (x in this case), then
factor by grouping . In the last step, use the difference of squares pattern to
complete
the factorization.
p(x) = x[x^{3} + 4x^{2} − 9x − 36]
p(x) = x[x^{2}(x + 4) − 9(x + 4)]
p(x) = x(x^{2} − 9)(x + 4)
p(x) = x(x + 3)(x − 3)(x + 4)
Set
0 = x(x + 3)(x − 3)(x + 4)
and use the zero product property to write
x = 0 or x + 3 = 0 or x − 3 = 0 or x + 4 = 0.
Solving, the zeros are x = 0, −3, 3, and −4.
21. Start with p(x) = −2x^{4} − 10x^{3} + 8x^{2} + 40x. Factor
out the gcf (−2x in this
case), then factor by grouping. In the last step , use the difference of squares
pattern
to complete the factorization.
p(x) = −2x[x^{3} + 5x^{2} − 4x − 20]
p(x) = −2x[x^{2}(x + 5) − 4(x + 5)]
p(x) = −2x(x^{2} − 4)(x + 5)
p(x) = −2x(x + 2)(x − 2)(x + 5)
Set
0 = −2x(x + 2)(x − 2)(x + 5)
and use the zero product property to write
−2x = 0 or x + 2 = 0 or x − 2 = 0 or x + 5 = 0.
Solving, the zeros are x = 0, −2, 2, and −5.
23. Start with p(x) = 2x^{3} − 7x^{2} − 15x. Factor out the
gcf (x in this case), then use
the ac method to complete the factorization.
p(x) = x[2x^{2} − 7x − 15]
p(x) = x[2x^{2} − 10x + 3x − 15]
p(x) = x[2x(x − 5) + 3(x − 5)]
p(x) = x(2x + 3)(x − 5)
Set
0 = x(2x + 3)(x − 5)
and use the zero product property to write
x = 0 or 2x + 3 = 0 or x − 5 = 0.
Solving, the zeros are x = 0, −3/2, and 5.
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