Pre-Calculus 12
Unit 3.3 Assignment.
Name: ______Key____________
1. What is the corresponding binomial factor of a polynomial, P(x), given the value of the zero?
a) P(1) = 0
c) P(4) = 0
If P(1) = 0, then a corresponding
If P(4) = 0, then a corresponding binomial
binomial factor of a polynomial, P(x), is x – 1.
b) P(−3) = 0
factor of a polynomial, P(x), is x – 4.
d) P(a) = 0
If P(–3) = 0, then a corresponding
If P(a) = 0, then a corresponding binomial
binomial factor of a polynomial, P(x), is x + 3.
factor of a polynomial, P(x), is x – a.
2. Determine whether x − 1 is a factor of each polynomial.
a) x3 − 3x2 + 4x – 2
Evaluate P(1).
P(1) = (1)3 – 3(1)2 + 4(1) – 2
= 0 Yes it is a factor.
b) 2x3 − x2 − 3x − 2
Evaluate P(1).
P(1) = 2(1)3 – (1)2 – 3 (1) – 2
= –4 No, it is not a factor.
d) 2x3 + 4x2 − 5x − 1
Evaluate P(1).
P(1) = 2(1)3 + 4(1) 2 – 5(1) – 1
=0
Yes it is a factor.
e) x4 − 3x3 + 2x2 − x + 1
Evaluate P(1).
P(1) = (1)4 – 3(1)3 + 2(1)2 – (1) + 1
=0
Yes it is a factor.
c) 3x3 − x − 3
Evaluate P(1).
P(1) = 3(1)3 – (1)2 – 3
= –1 No, it is not a factor.
3. State whether each polynomial has x + 2 as a factor.
a) 5x2 + 2x + 6
Evaluate P(–2).
P(–2) = 5(–2)2 + 2(–2) + 6
= 22 No, it is not a factor.
b) 2x3 − x2 − 5x − 8
Evaluate P(–2).
P(–2) = 2(–2) 3 – (–2)2 – 5(–2) –8
= –18 No, it is not a factor.
c) 2x3 + 2x2 − x − 6
Evaluate P(–2).
P(–2) = 2(–2)3 + 2(–2)2 – (–2) – 6
= –12 No, it is not a factor.
d) x4 − 2x2 + 3x − 4
Evaluate P(–2).
P(–2) = (–2)4 – 2(–2)2 + 3(–2) – 4
= –2 No, it is not a factor.
4. What are the possible integral zeros of each polynomial?
a) P(x) = x3 + 3x2 − 6x − 8
c) P(n) = n3 − 3n2 − 10n + 24
The possible integral zeros of the
polynomial are the factors of the
constant term, –8: ±1, ±2, ±4,
and ±8.
b) P(s) = s3 + 4s2 − 15s − 18
The possible integral zeros of the
polynomial are the factors of the
constant term, 24: ±1, ±2, ±3, ±4, ±6,
±8, ±12 and ±24.
d) P(p) = p4 − 2p3 − 8p2 + 3p – 4
The possible integral zeros of the
polynomial are the factors of the
constant term, –18: ±1, ±2, ±3,
±6, ±9 and ±18.
The possible integral zeros of the
polynomial are the factors of the
constant term, –4: ±1, ±2, ±3, ±4,
5. Factor fully.
a) P(x) = x3 − 6x2 + 11x – 6
Use synthetic division, and try –1.
b) P(x) = x3 + 2x2 − x – 2
c) P(v) = v3 + v2 − 16v – 16
d) P(x) = x4 + 4x3 − 7x2 − 34x – 24
Use synthetic division, and try –1.
Use synthetic division, and try +1.
Use synthetic division, and try +1.
6. Determine the value(s) of k so that the binomial is a factor of the polynomial.
a) x2 − x + k, x − 2
b) x2 − 6x − 7, x + k
c) x3 + 4x2 + x + k, x + 2
d) x2 + kx − 16, x − 2
7. The volume, V(h), of a bookcase can be represented by the expression h3 − 2h2 + h, where h is the
height of the bookcase. What are the possible dimensions of the bookcase in terms of h?
8. When a certain type of plastic is cut into sections, the length of each section determines its relative
strength. The function f (x) = x4 − 14x3 + 69x2 − 140x + 100 describes the relative strength of a
section of length x feet. After testing the plastic, engineers discovered that 5-ft sections were
extremely weak.
a) Why is x − 5 a possible factor when x = 5 is the length of the pipe? Show that x − 5 is a factor of
the polynomial function. x – 5 is a possible factor because it is the corresponding factor for x = 5.
b) Are there other lengths of plastic that are extremely weak?
9. The product of four integers is x4 + 6x3 + 11x2 + 6x, where x is one of the integers. What are the
possible expressions for the other three integers?
10. Determine the values of m and n so that the polynomials 2x3 + mx2 + nx − 3 and x3 − 3mx2 + 2nx + 4
are both divisible by x − 2.