Multiply & Divide Rational Expressions

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Question 1 of 6

1. Question

Multiply

$(3x^2)/(4x) xx (5x^2)/(10x^3)$

1. $5/8$
2. $40$
3. $(3x^8)/8$
4. $3/8$

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Multiplying Rational Expressions

$a/b xx c/d = (ab)/(cd)$

Multiply the terms in the numerator and then in the denominator.

$(3x^2)/(4x) xx (5x^2)/(10x^3)$	$=$	$(3x^2 xx 5x^2)/(4x xx 10x^3)$

	$=$	$((3xx5)x^(2+2))/((4xx10) x^(1+3)$	Multiply the constants and the variables

	$=$	$(15x^4)/(40x^4)$	Apply Multiplication Rule of Exponents

	$=$	$15/40$
	$=$	$3/8$	Express in lowest terms

$3/8$

Question 2 of 6

2. Question

Multiply

$(m^2-5m-6)/(m-2) xx (4m-8)/(m-6)$

1. $4(m-1)$
2. $4$
3. $4(m+1)$
4. $m+1$

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Multiplying Rational Expressions

$a/b xx c/d = (ab)/(cd)$

Look for polynomials that can be factorised before proceeding with the operation. Since the numerator of the first term is in standard form

$($ $a$

$x^2+$ $b$

$x+$ $c$

$=0)$ we can factorise using the cross method.

$m^2$ $-5$

$m$ $-6$

$=0$

To factorise, we need to find two numbers that add to $-5$ and multiply to $-6$

$-6$ and

$1$ fit both conditions

$-6+1$	$=$	$-5$
$-6 xx 1$	$=$	$-6$

Read across to get the factors.

$(m-6)(m+1)=0$

Do the same for the numerator of the second term.

$4m-8$

$=$

$4(m-2)$

Rewrite the expression with the factors.

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

	$=$	$(m+1) xx 4$	Cancel out $m-6$ and $m-2$
	$=$	$4(m+1)$	Simplify

$4(m+1)$

Question 3 of 6

3. Question

Multiply

$(4x^2+4x)/(5x^2-5x-60) xx (x^2-6x+8)/(x^2-2x)$

1. $(4(x+3))/(5(x-2))$
2. $(4(x+1))/(5(x+3))$
3. $(4(x+2))/(5(x+3))$
4. $4/5$

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Multiplying Rational Expressions

$a/b xx c/d = (ab)/(cd)$

Look for polynomials that can be factorised before proceeding with the operation.

$4x^2+4x$	$=$	$4x(x+1)$
$5x^2-5x-60$	$=$	$5(x^2-x-12)$
$x^2-2x$	$=$	$x(x-2)$

The denominator of the first term is in standard form

$($ $a$

$x^2+$ $b$

$x+$ $c$

$=0)$ so we can factorise using the cross method.

$x^2$ $-$

$x$ $-12$

$=0$

To factorise, we need to find two numbers that add to $-1$ and multiply to $-12$

$-4$ and

$3$ fit both conditions

$-4+3$	$=$	$-1$
$-4 xx 3$	$=$	$-12$

Read across to get the factors.

$(x-4)(x+3)=0$

The numerator of the second term is also in standard form

$($ $a$

$x^2+$ $b$

$x+$ $c$

$=0)$ so we can factorise using the cross method.

$x^2$ $-6$

$x+$ $8$

$=0$

To factorise, we need to find two numbers that add to $-6$ and multiply to $8$

$-4$ and

$-2$ fit both conditions

$-4-2$	$=$	$-6$
$-4 xx -2$	$=$	$8$

Read across to get the factors.

$(x-4)(x-2)=0$

Rewrite the expression with the factors.

$(4x^2+4x)/(5x^2-5x-60) xx (x^2-6x+8)/(x^2-2x)$	$=$	$(4x(x+1))/(5(x-4)(x+3)) xx ((x-4)(x-2))/(x(x-2))$

	$=$	$(4(x+1))/(5(x+3))$	Cancel out $x$ , $x-4$ and $x-2$

$(4(x+1))/(5(x+3))$

Question 4 of 6

4. Question

Divide

$(6a^2+11a-10)/(3a-2) -: (2a+5)$

1. $2a+5$
2. $3a-2$
3. $2$
4. $1$

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Dividing Rational Expressions

$a/b -: c/d = a/b xx d/c$

Since the numerator of the first term is in standard form

$($ $a$

$x^2+$ $b$

$x+$ $c$

$=0)$ we can factorise using the cross method.

$6$

$a^2+$ $11$

$a$ $-10$

$=0$

To factorise, we need to find two numbers that add to $11$ and multiply to $-10$

$3a$ ,

$2a$ ,

$-2$ and

$5$ fit both conditions

$15a – 4a$	$=$	$11$ $a$
$-2 xx 5$	$=$	$-10$

Read across to get the factors.

$(3a-2)(2a+5)=0$

Rewrite the expressions

$(6a^2+11a-10)/(3a-2) -: (2a+5)$	$=$	$((3a-2)(2a+5))/(3a-2) -: (2a+5)$

	$=$	$((3a-2)(2a+5))/(3a-2) xx 1/(2a+5)$	Apply Division Formula

	$=$	$1$	$3a-2$ and $2a+5$ cancel out in the denominator

$1$

Question 5 of 6

5. Question

Divide

$(3x^2-9x)/(x^2-12x+36) -: (x^3-9x)/(x-6)$

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

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Dividing Rational Expressions

$a/b -: c/d = a/b xx d/c$

Since the denominator of the first term is in standard form

$($ $a$

$x^2+$ $b$

$x+$ $c$

$=0)$ we can factorise using the cross method.

$x^2$ $-12$

$x+$ $36$

$=0$

To factorise, we need to find two numbers that add to $-12$ and multiply to $36$

$-6$ and

$-6$ fit both conditions

$-6 – 6$	$=$	$-12$
$-6 xx -6$	$=$	$36$

Read across to get the factors.

$(x-6)(x-6)=0$

Factorise remaining expressions.

$3x^2-9x$	$=$	$3x(x-3)$	Factor $3x$ out
$x^3-9x$	$=$	$x(x^2-9)$	Factor $x$ out
	$=$	$x(x-3)(x+3)$	$x^2-9 = (x-3)(x+3)$

Rewrite the expressions

$(3x^2-9x)/(x^2-12x+36) -: (x^3-9x)/(x-6)$	$=$	$(3x^2-9x)/(x^2-12x+36) xx (x-6)/(x^3-9x)$	Apply Division Formula

	$=$	$(3x(x-3))/((x-6)(x-6)) xx (x-6)/(x(x-3)(x+3))$	Show factors

	$=$	$3/((x-6)(x+3))$	Cancel out like terms

$3/((x-6)(x+3))$

Question 6 of 6

6. Question

Divide

$(m^2+2m-8)/(m^2-3m+2) -: (m^2+5m+4)/(m^2-4m+3)$

1. $(m-3)/(m+1)$
2. $(m+3)/(m+1)$
3. $m+3$
4. $(m+1)/(m-3)$

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Dividing Rational Expressions

$a/b -: c/d = a/b xx d/c$

Since the given polynomials are in standard form

$($ $a$

$x^2+$ $b$

$x+$ $c$

$=0)$ we can factorise using the cross method.

$m^2+$ $2$

$m$ $-8$

$=0$

To factorise, we need to find two numbers that add to $2$ and multiply to $-8$

$4$ and

$-2$ fit both conditions

$4 – 2$	$=$	$2$
$4 xx -2$	$=$	$-8$

Read across to get the factors.

$(m+4)(m-2)=0$

Do the same for the remaining polynomials.

$m^2$ $-3$

$m+$ $2$

$=0$

To factorise, we need to find two numbers that add to $-3$ and multiply to $2$

$-2$ and

$-1$ fit both conditions

$-2 – 1$	$=$	$-3$
$-2 xx -1$	$=$	$2$

Read across to get the factors.

$(m-2)(m-1)=0$

$m^2+$ $5$

$m+$ $4$

$=0$

To factorise, we need to find two numbers that add to $5$ and multiply to $4$

$4$ and

$1$ fit both conditions

$4 + 1$	$=$	$5$
$4 xx 1$	$=$	$4$

Read across to get the factors.

$(m+4)(m+1)=0$

$m^2$ $-4$

$m+$ $3$

$=0$

To factorise, we need to find two numbers that add to $-4$ and multiply to $3$

$-3$ and

$-1$ fit both conditions

$-3 – 1$	$=$	$-4$
$-3 xx -1$	$=$	$3$

Read across to get the factors.

$(m-3)(m-1)=0$

Rewrite the expressions

$(m^2+2m-8)/(m^2-3m+2) -: (m^2+5m+4)/(m^2-4m+3)$	$=$	$(m^2+2m-8)/(m^2-3m+2) xx (m^2-4m+3)/(m^2+5m+4)$	Apply Division Formula

	$=$	$((m+4)(m-2))/((m-2)(m-1)) xx ((m-3)(m-1))/((m+4)(m+1))$	Show factors

	$=$	$(m-3)/(m+1)$	Cancel out like terms

$(m-3)/(m+1)$

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