Remainder Theorem

Years

Try VividMath Premium to unlock full access

Start Free Trial

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

Question 1 of 4

Write “F” if the binomial is a Factor and “N” if it is not a Factor of the polynomial:

P (x) = x^{3} + 2 x^{2} - 5 x - 6

$P(x)=x^3+2x^2-5x-6$

$(a) (x - 2) =$ $(a) (x-2)=$ (F, f)

$(b) (2 x - 1) =$ $(b) (2x-1)=$ (N, n)

$(c) (x + 3) =$ $(c) (x+3)=$ (F, f)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

\frac{P (x)}{x - a}

$(P(x))/(x-a)$ then

P (a) = r

$P(a)=r$

where

r

$r$ is the Remainder

Note that if

r = 0

$r=0$ , then

(x - a)

$(x-a)$ is a Factor of

P (x)

$P(x)$

(a)

$(a)$ Identify if

(x - 2)

$(x-2)$ is a factor of

P (x) = x^{3} + 2 x^{2} - 5 x - 6

$P(x)=x^3+2x^2-5x-6$ .

First, label the variable

a

$a$ by equating the binomial to

0

$0$ and solving for

x

$x$

$x - 2$ $x-2$	$=$ $=$	$0$ $0$
$x - 2$ $x-2$ $+ 2$ $+2$	$=$ $=$	$0$ $0$ $+ 2$ $+2$	Add $2$ $2$ to both sides
$x$ $x$	$=$ $=$	$2$ $2$

Hence,

a = 2

$a=2$

Next, substitute

a

$a$ into the polynomial to solve for

r

$r$

P (x) = x^{3} + 2 x^{2} - 5 x - 6

$P(x)=x^3+2x^2-5x-6$

a = 2

$a=2$

$P (a)$ $P(a)$	$=$ $=$	$r$ $r$	Remainder Theorem
$P (a)$ $P(a)$	$=$ $=$	$a^{3} + 2 a^{2} - 5 a - 6$ $a^3+2a^2-5a-6$	Replace $x$ $x$ with $a$ $a$
$P (2)$ $P(2)$	$=$ $=$	$2^{3} + 2 {(2)}^{2} - 5 (2) - 6$ $2^3+2(2)^2-5(2)-6$	Substitute $a = 2$ $a=2$
	$=$ $=$	$8 + 2 (4) - 10 - 6$ $8+2(4)-10-6$
	$=$ $=$	$8 + 8 - 10 - 6$ $8+8-10-6$
	$=$ $=$	$16 - 16$ $16-16$
$r$ $r$	$=$ $=$	$0$ $0$

Since

r = 0

$r=0$ , we can say that

(x - 2)

$(x-2)$ is a factor of

P (x) = x^{3} + 2 x^{2} - 5 x - 6

$P(x)=x^3+2x^2-5x-6$

(b)

$(b)$ Identify if

(2 x - 1)

$(2x-1)$ is a factor of

P (x) = x^{3} + 2 x^{2} - 5 x - 6

$P(x)=x^3+2x^2-5x-6$ .

First, label the variable

$a$ by equating the binomial to

$0$ and solving for

$x$

$2x-1$	$=$	$0$
$2x-1$ $+1$	$=$	$0$ $+1$	Add $1$ to both sides
$2x$	$=$	$1$
$2x$ $-:2$	$=$	$1$ $-:2$	Divide both sides by $2$

$x$	$=$	$1/2$

Hence,

$a=1/2$

Next, substitute

$a$ into the polynomial to solve for

$r$

$P(x)=x^3+2x^2-5x-6$

$a=1/2$

$P(a)$	$=$	$r$	Remainder Theorem
$P(a)$	$=$	$a^3+2a^2-5a-6$	Replace $x$ with $a$

$P(1/2)$	$=$	$(1/2)^3+2(1/2)^2-5(1/2)-6$	Substitute $a=1/2$

	$=$	$1/8+2(1/4)-5/2-6$

	$=$	$1/8+4/8-20/8-48/8$

$r$	$=$	$(-63)/8$

Since

$r\neq0$ , we can say that

$(2x-1)$ is not a factor of

$P(x)=x^3+2x^2-5x-6$

$(c)$ Identify if

$(x+3)$ is a factor of

$P(x)=x^3+2x^2-5x-6$ .

First, label the variable

$a$ by equating the binomial to

$0$ and solving for

$x$

$x+3$	$=$	$0$
$x+3$ $-3$	$=$	$0$ $-3$	Subtract $3$ from both sides
$x$	$=$	$-3$

Hence,

$a=-3$

Next, substitute

$a$ into the polynomial to solve for

$r$

$P(x)=x^3+2x^2-5x-6$

$a=-3$

$P(a)$	$=$	$r$	Remainder Theorem
$P(a)$	$=$	$a^3+2a^2-5a-6$	Replace $x$ with $a$

$P(-3)$	$=$	$(-3)^3+2(-3)^2-5(-3)-6$	Substitute $a=-3$

	$=$	$-27+2(9)+15-6$
	$=$	$-27+18+15-6$
	$=$	$-33+33$
$r$	$=$	$0$

Since

$r=0$ , we can say that

$(x+3)$ is a factor of

$P(x)=x^3+2x^2-5x-6$

$(a)$ Factor

$(b)$ Not a factor

$(c)$ Factor

Question 2 of 4

2. Question
Solve for the remainder of:

$(8x+x^2+3)divide(3+x)$
- $r=$ (-12)
Hint

Help Video
Correct

Well Done!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

Remainder Theorem

If $(P(x))/(x-a)$ then $P(a)=r$

where $r$ is the Remainder

Long Division

Use long division when a polynomial is divided by a binomial

Note that if $r=0$ , then $(x-a)$ is a Factor of $P(x)$

Method One

Find the remainder using long division

First, arrange the terms in descending order of powers.

$\mathsf{P}$ (Polynomial) $=$ $8x+x^2+3$

$=$ $x^2+8x+3$

$\mathsf{Divisor}$ $=$ $3+x$

$=$ $x+3$

< $=$

Next, solve for each term of the quotient

First term of the quotient:

Divide the first term of the Polynomial by the first term of the Divisor. Place this above the Polynomial

$x^2dividex$ $=$ $x$

Multiply $x$ to the divisor. Place this under the Polynomial

$x$ $(x+3)$ $=$ $x^2+3x$

Subtract $x^2+3x$ and write the difference one line below

Drop down $3$ and repeat the process to get the second term of the quotient

Second term of the quotient:

Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial

$5xdividex$ $=$ $5$

Multiply $5$ to the divisor. Place this under the Polynomial

$5$ $(x+3)$ $=$ $5x+15$

Subtract $5x+15$ and write the difference one line below

$-12$ is not divisible by the divisor ( $x+3$ ) anymore, hence $r=-12$

$r=-12$

Method Two

Find the remainder using remainder theorem

First, label the variable $a$ by equating the divisor to $0$ and solving for $x$

$x+3$ $=$ $0$

$x+3$ $-3$ $=$ $0$ $-3$ Subtract $3$ from both sides

$x$ $=$ $-3$

Hence, $a=-3$

Next, substitute $a$ into the polynomial to solve for $r$

$P(x)=8x+x^2+3$

$a=-3$

$P(a)$ $=$ $r$ Remainder Theorem

$P(a)$ $=$ $8a+a^2+3$ Replace $x$ with $a$

$P(-3)$ $=$ $8(-3)+(-3)^2+3$ Substitute $a=-3$

$=$ $-24+9+3$

$r$ $=$ $-12$

$r=-12$

Question 3 of 4

Solve for the remainder of:

$(x^4-3x^3+2x+8)/(x-2)$

$r=$ (4)

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Remainder Theorem

$(P(x))/(x-a)$ then

$P(a)=r$

where

$r$ is the Remainder

Long Division

Use long division when a polynomial is divided by a binomial

Note that if

$r=0$ , then

$(x-a)$ is a Factor of

$P(x)$

Method One

Find the remainder using long division

First, notice that the term with the power of

$2$ is missing. Add this term to the polynomial before using long division

$\mathsf{P}$ (Polynomial)	$=$	$x^4-3x^3+2x+8$
	$=$	$x^4-3x^3+$ $0x^2$ $+2x+8$
$\mathsf{Divisor}$	$=$	$x-2$

$=$

Next, solve for each term of the quotient

First term of the quotient:

Divide the first term of the Polynomial by the first term of the Divisor. Place this above the Polynomial

$x^4dividex$

$=$

$x^3$

Multiply $x^3$ to the divisor. Place this under the Polynomial

$x^3$

$(x-2)$

$=$

$x^4-2x^3$

Subtract $x^4-2x^3$ and write the difference one line below

Drop down

$0x^2$ and repeat the process to get the second term of the quotient

Second term of the quotient:

Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial

$-x^3dividex$

$=$

$-x^2$

Multiply $-x^2$ to the divisor. Place this one line below

$-x^2$

$(x-2)$

$=$

$-x^3+2x^2$

Subtract $-x^3+2x^2$ and write the difference one line below

Drop down

$2x$ and repeat the process to get the third term of the quotient

Third term of the quotient:

Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial

$-2x^2dividex$

$=$

$-2x$

Multiply $-2x$ to the divisor. Place this under the Polynomial

$-2x$

$(x-2)$

$=$

$-2x^2+4x$

Subtract $-2x^2+4x$ and write the difference one line below

Drop down

$8$ and repeat the process to get the fourth term of the quotient

Fourth term of the quotient:

Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial

$-2xdividex$

$=$

$-2$

Multiply $-2$ to the divisor. Place this under the Polynomial

$-2$

$(x-2)$

$=$

$-2x+4$

Subtract $-2x+4$ and write the difference one line below

$4$ is not divisible by the divisor (

$x-2$ ) anymore, hence

$r=4$

Method Two

Find the remainder using remainder theorem

First, label the variable

$a$ by equating the divisor to

$0$ and solving for

$x$

$x-2$	$=$	$0$
$x-2$ $+2$	$=$	$0$ $+2$	Add $2$ to both sides
$x$	$=$	$2$

Hence,

$a=2$

Next, substitute

$a$ into the polynomial to solve for

$r$

$P(x)=x^4-3x^3+2x+8$

$a=2$

$P(a)$	$=$	$r$	Remainder Theorem
$P(a)$	$=$	$a^4-3a^3+2a+8$	Replace $x$ with $a$
$P(2)$	$=$	$2^4-3(2)^3+2(2)+8$	Substitute $a=2$
	$=$	$16-3(8)+4+8$
	$=$	$16-24+4+8$
$r$	$=$	$4$

$r=4$

Question 4 of 4

Solve for the remainder of:

$(a) (x^4-5x^3+3x^2+8)/(x-2)$

$(b) (3x^4-x^2+2x+6)/(x+1)$

$(a) r=$ (-4)

$(b) r=$ (6)

Remainder Theorem

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Correct!

Remainder Theorem

2. Question

Hint

Well Done!

Remainder Theorem

Long Division

3. Question

Hint

Excellent!

Remainder Theorem

Long Division

4. Question

Hint

Great Work!

Remainder Theorem

Quizzes

$\mathsf{P}$ (Polynomial)	$=$	$8x+x^2+3$
	$=$	$x^2+8x+3$
$\mathsf{Divisor}$	$=$	$3+x$
	$=$	$x+3$

$x+1$	$=$	$0$
$x+1$ $-1$	$=$	$0$ $-1$	Subtract $1$ from both sides
$x$	$=$	$-1$