POLYNOMIAL FUNCTIONS
Test your understanding of polynomial functions, including identifying degrees, leading coefficients, end behavior, factoring, graph interpretation, and solving polynomial equations.
1 / 55
When dividing
x3−2x2+3x−4
by
x−1
, the quotient is:
Use polynomial long division or synthetic division to find the quotient.
2 / 55
The remainder when 2x3−5x2+4x−1 is divided by x−2 is:
Substitute x=2 into the polynomial: 2(8)−5(4)+4(2)−1=3.
3 / 55
If
x4−3x3+2x2−x+5
is divided by
x2−1
, the remainder is:
Perform polynomial long division; the remainder is linear.
4 / 55
Synthetic division is used to divide a polynomial by:
Synthetic division simplifies division by linear terms.
5 / 55
The quotient when
x3−6x2+11x−6
x−3
is:
Use synthetic division or long division to verify.
6 / 55
When dividing 3x4−4x3+2x2−5x+1 by x+1, the remainder is:
Substitute x=−1: 3(1)−4(−1)+2(1)−5(−1)+1=15.
7 / 55
The degree of the quotient when a 4th-degree polynomial is divided by a 2nd-degree polynomial is:
Degree of quotient = Degree of dividend – Degree of divisor.
8 / 55
x3+2x2−5x−6
x+3
Use synthetic division to confirm.
9 / 55
The remainder when x5−1 is divided by x−1 is:
x−1 is a factor, so remainder is 0.
10 / 55
The division algorithm states that for polynomials
P(x)
and
D(x)≠0
, there exist unique polynomials
Q(x)
R(x)
such that:
This is the standard division algorithm for polynomials.
11 / 55
The Remainder Theorem states that the remainder when
x−c
Direct application of the Remainder Theorem.
12 / 55
If P(x)=2x3−3x2+4x−1, then P(1) equals:
P(1)=2(1)−3(1)+4(1)−1=2.
13 / 55
The remainder when x4−2x3+x2−3x+5 is divided by x−2 is:
Substitute x=2: 16−16+4−6+5=7.
14 / 55
leaves a remainder of 5 when divided by
, then:
By the Remainder Theorem.
15 / 55
The remainder when 3x3−4x2+5x−2 is divided by x+1 is:
Substitute x=−1: −3−4−5−2=−14.
16 / 55
If P(x)=x3+kx2−4x+3 and P(2)=7, then k is:
8+4k−8+3=7⇒k=1.
17 / 55
The remainder when x5+2x3−x2+4 is divided by x−1 is:
Substitute x=1: 1+2−1+4=6.
18 / 55
and the remainder is 0, then:
Factor Theorem.
19 / 55
The remainder when
4x3−3x2+2x−1
x−12
Substitute
x=12
:
4(18)−3(14)+2(12)−1=−12
.
20 / 55
If P(x)=x3−4x2+5x−2, then P(2) is:
P(2)=8−16+10−2=0.
21 / 55
The Factor Theorem states that
is a factor of
if and only if:
Direct application of the Factor Theorem.
22 / 55
Which of the following is a factor of
x3−3x2−4x+12
?
P(2)=8−12−8+12=0
23 / 55
x+1
P(x)=x3+2x2−x−2
, then another factor is:
Factor as
(x+1)(x2+x−2)=(x+1)(x−1)(x+2)
24 / 55
The polynomial
P(x)=x3−6x2+11x−6
has factors:
Verify by substitution or synthetic division.
25 / 55
If x−3 is a factor of x3−kx2+5x−6, then k is:
P(3)=27−9k+15−6=0⇒k=4.
26 / 55
The number of distinct real roots of P(x)=x3−3x2+2x is:
Factor as x(x−1)(x−2); roots at x=0,1,2.
27 / 55
P(x)=x4−5x2+4
, then one of its factors is:
P(1)=1−5+4=0
28 / 55
P(x)=x3+x2−4x−4
has a factor:
P(−2)=−8+4+8−4=0
29 / 55
x−2
P(x)=x2−2
x2−2=(x−2)(x+2)
30 / 55
P(x)=x3−2x2−x+2
has roots:
(x−1)(x+1)(x−2)
31 / 55
The end behavior of P(x)=−x3+2x2−x+1 is:
Odd degree, negative leading coefficient.
32 / 55
The number of turning points of P(x)=x4−3x2+2 is at most:
A degree-n polynomial has at most n−1 turning points.
33 / 55
The graph of
P(x)=(x−1)2(x+2)
crosses the x-axis at:
Odd multiplicity crosses, even multiplicity touches.
34 / 55
The y-intercept of
P(x)=2x3−4x2+x−5
x=0
35 / 55
P(x)=x3−4x
has zeros at:
x(x2−4)
36 / 55
The graph of P(x)=x2(x−3) resembles:
Odd degree, positive leading term after expansion.
37 / 55
The multiplicity of the root x=2 in P(x)=(x−2)3(x+1) is:
The exponent of the factor (x−2).
38 / 55
The graph of P(x)=x4−1 has symmetry about:
Even function: P(−x)=P(x).
39 / 55
The number of real roots of P(x)=x3+x2+x+1 is:
Only x=−1 is a real root; others are complex.
40 / 55
has x-intercepts at:
(x−1)(x−2)(x−3)
41 / 55
The solution to
(x−1)(x+2)(x−3)>0
Test intervals defined by roots
x=−2,1,3
42 / 55
The inequality
x2−4x+3≤0
holds for:
(x−1)(x−3)
; parabola opens upwards.
43 / 55
x3−4x2+x+6≥0
includes:
(x+1)(x−2)(x−3)
and test intervals.
44 / 55
x(x−2)(x+1)<0
is true for:
x=−1,0,2
45 / 55
x4−5x2+4≤0
(x2−1)(x2−4)
and solve.
46 / 55
x3+2x2−x−2>0
(x+2)(x+1)(x−1)
47 / 55
(x+3)(x−1)2≥0
The squared term ensures non-negativity at
x=1
48 / 55
x3−8<0
(x−2)(x2+2x+4)
; only
x=2
is a real root.
49 / 55
x4−16≥0
(x2−4)(x2+4)
50 / 55
(x−1)3(x+2)2≤0
The squared term is always non-negative; the cubic term changes sign at
51 / 55
The partial fraction decomposition of
3x+2(x−1)(x+2)
Factor denominator and solve for A and B.
52 / 55
The partial fractions of
x2+1x(x−1)2
include terms:
Repeated linear factors require multiple terms.
53 / 55
5x−1x2−x−2
54 / 55
2x+3(x+1)(x2+1)
include:
An irreducible quadratic factor requires a linear numerator.
55 / 55
x2+2x−1x3−x
Factor denominator and solve for A, B, C.
Your score is
Restart quiz Exit