Finding the Lowest Common Denominator

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Question 1 of 6

1. Question
Find the lowest common denominator (LCD) of

$6 x$ $6x$ and $16 x^{2}$ $16x^2$
- 1. $12 x^{2}$ $12x^2$
- 2. $48 x^{2}$ $48x^2$
- 3. $12$ $12$
- 4. $48 x$ $48x$
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The lowest common denominator (LCD) can also be referred to as the least common multiple (LCM).

First, decompose each term into its smallest factors.

$6 x$ $6x$ $:$ $:$ $2$ $2$ , $3$ $3$ , $x$ $x$

$16 x^{2}$ $16x^2$ $:$ $:$ $2$ $2$ , $2, 2, 2,$ $2,2,2,$ $x$ $x$ , $x$ $x$

Combine the red terms and multiply to the black ones.

$LCD$ $\text(LCD)$ $:$ $:$ $2$ $2$ , $x$ $x$ , $3, 2, 2, 2, x$ $3,2,2,2,x$

$:$ $:$ $2$ $2$ $\times$ $xx$ $x$ $x$ $\times 3 \times 2 \times 2 \times 2 \times x$ $xx 3 xx 2 xx 2 xx 2 xx x$ Multiply the factors

$:$ $:$ $48 x^{2}$ $48x^2$

$48 x^{2}$ $48x^2$
Question 2 of 6

2. Question
Find the lowest common denominator (LCD) of

$x - 3$ $x-3$ and $3 x + 5$ $3x+5$
- 1. $(x - 3) (3 x + 5)$ $(x-3)(3x+5)$
- 2. $x - 3$ $x-3$
- 3. $3 x + 5$ $3x+5$
- 4. $x + 5$ $x+5$
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The lowest common denominator (LCD) can also be referred to as the least common multiple (LCM).

First, look for the factors of the given terms.

$x - 3$ $x-3$ $:$ $:$ $x - 3, 1$ $x-3, 1$

$3 x - 5$ $3x-5$ $:$ $:$ $3 x - 5, 1$ $3x-5, 1$

Since $1$ $1$ is the only common factor, then we can find the LCD.

$LCD$ $\text(LCD)$ $:$ $:$ $(x - 3) (3 x + 5)$ $(x-3)(3x+5)$

$(x - 3) (3 x + 5)$ $(x-3)(3x+5)$
Question 3 of 6

3. Question
Find the lowest common denominator (LCD) of

${(x - 2)}^{2}$ $(x-2)^2$ and $x^{2} + 2 x + 8$ $x^2+2x+8$
- 1. $(x + 4) (x + 2)$ $(x+4)(x+2)$
- 2. $(x - 2) (x + 4)$ $(x-2)(x+4)$
- 3. $(x + 2) (x - 2) (x + 4)$ $(x+2)(x-2)(x+4)$
- 4. $(x - 2) (x - 2) (x + 4)$ $(x-2)(x-2)(x+4)$
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The lowest common denominator (LCD) can also be referred to as the least common multiple (LCM).

First, decompose each term into its smallest factors.

${(x - 2)}^{2}$ $(x-2)^2$ $=$ $=$ $(x - 2) \times (x - 2)$ $(x-2) xx (x-2)$

Since the second term is in standard form $($ $($ $a$ $a$ $x^{2} +$ $x^2+$ $b$ $b$ $x +$ $x+$ $c$ $c$ $= 0)$ $=0)$ we can factorise using the cross method.

$x^{2} +$ $x^2+$ $2$ $2$ $x$ $x$ $- 8$ $-8$ $= 0$ $=0$

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

$4$ $4$ and $- 2$ $-2$ fit both conditions

$4 - 2$ $4 – 2$ $=$ $=$ $2$ $2$

$4 \times - 2$ $4 xx -2$ $=$ $=$ $- 8$ $-8$

Read across to get the factors.

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

Write the factors of the two polynomials next to each other.

$(x-2)^2$ $=$ $(x-2)$ $(x-2)$

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

Combine the red terms and multiply to the black ones.

$\text(LCD)$ $=$ $(x-2)$ $(x-2)(x+4)$

$(x-2)(x-2)(x+4)$
Question 4 of 6

4. Question
Find the lowest common denominator (LCD) of

$x^2-5x+6$ and $x^2-x-6$
- 1. $(x-3)(x+3)(x+2)$
- 2. $(x-3)(x-2)(x+2)$
- 3. $(x-3)(x-2)(x-2)$
- 4. $(x+3)(x-2)(x+2)$
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The lowest common denominator (LCD) can also be referred to as the least common multiple (LCM).

Since the first polynomial is in standard form $($ $a$ $x^2+$ $b$ $x+$ $c$ $=0)$ we can factorise using the cross method.

$x^2$ $-5$ $x+$ $6$ $=0$

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

$-3$ and $-2$ fit both conditions

$-3 – 2$ $=$ $-5$

$-3 xx -2$ $=$ $6$

Read across to get the factors.

$(x-3)(x-2)=0$

Since the first polynomial is in standard form $($ $a$ $x^2+$ $b$ $x+$ $c$ $=0)$ we can factorise using the cross method.

$x^2$ $-$ $x$ $-6$ $=0$

To factorise, we need to find two numbers that add to $-1$ and multiply to $-6$

$-3$ and $2$ fit both conditions

$-3 + 2$ $=$ $-1$

$-3 xx 2$ $=$ $-6$

Read across to get the factors.

$(x-3)(x+2)=0$

Write the factors of the two polynomials next to each other.

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

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

Combine the red terms and multiply to the black ones.

$\text(LCD)$ $=$ $(x-3)$ $(x-2)(x+2)$

$(x-3)(x-2)(x+2)$
Question 5 of 6

5. Question
Find the lowest common denominator (LCD) of

$x+4$ and $x^2+8x+16$
- 1. $x+4$
- 2. $(x+4)(x-4)$
- 3. $x-4$
- 4. $(x+4)(x+4)$
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The lowest common denominator (LCD) can also be referred to as the least common multiple (LCM).

Since the second polynomial is in standard form $($ $a$ $x^2+$ $b$ $x+$ $c$ $=0)$ we can factorise using the cross method.

$x^2+$ $8$ $x+$ $16$ $=0$

To factorise, we need to find two numbers that add to $8$ and multiply to $16$

$4$ and $4$ fit both conditions

$4 + 4$ $=$ $8$

$4 xx 4$ $=$ $16$

Read across to get the factors.

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

Write the factors of the two polynomials next to each other and mark similar factors.

$x+4$ $=$ $(x+4)$ $(1)$

$x^2+8x+16$ $=$ $(x+4)$ $(x+4)$

Combine the red terms and multiply to the black ones.

$\text(LCD)$ $=$ $(x+4)$ $(x+4)$

$(x+4)(x+4)$
Question 6 of 6

6. Question
Find the lowest common denominator (LCD) of

$x$ , $x+2$ and $x^2-4$
- 1. $x(x+2)$
- 2. $(x+2)(x-2)$
- 3. $x(x-2)(x-2)$
- 4. $x(x+2)(x-2)$
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The lowest common denominator (LCD) can also be referred to as the least common multiple (LCM).

First, decompose each term into its smallest factors.

$x$ $=$ $x xx 1$

$x+2$ $=$ $(x+2) xx 1$

$x^2-4$ $=$ $(x-2)(x+2)$

Write the factors of the two polynomials next to each other and mark similar factors.

$x$ $:$ $x,1$

$x+2$ $:$ $(x-2)$ , $1$

$x^2-4$ $:$ $(x-2)$ , $(x+2)$

Combine the red terms and multiply to the black ones.

$\text(LCD)$ $=$ $(x-2)$ $x(x+2)$

$=$ $x(x+2)(x-2)$

$x(x+2)(x-2)$

Finding the Lowest Common Denominator

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Quiz summary

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1. Question

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2. Question

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3. Question

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4. Question

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Fantastic!

5. Question

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6. Question

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Quizzes

$6 x$ $6x$	$:$ $:$	$2$ $2$ , $3$ $3$ , $x$ $x$
$16 x^{2}$ $16x^2$	$:$ $:$	$2$ $2$ , $2, 2, 2,$ $2,2,2,$ $x$ $x$ , $x$ $x$

$LCD$ $\text(LCD)$	$:$ $:$	$2$ $2$ , $x$ $x$ , $3, 2, 2, 2, x$ $3,2,2,2,x$
	$:$ $:$	$2$ $2$ $\times$ $xx$ $x$ $x$ $\times 3 \times 2 \times 2 \times 2 \times x$ $xx 3 xx 2 xx 2 xx 2 xx x$	Multiply the factors
	$:$ $:$	$48 x^{2}$ $48x^2$

$x - 3$ $x-3$	$:$ $:$	$x - 3, 1$ $x-3, 1$
$3 x - 5$ $3x-5$	$:$ $:$	$3 x - 5, 1$ $3x-5, 1$

$4 - 2$ $4 – 2$	$=$ $=$	$2$ $2$
$4 \times - 2$ $4 xx -2$	$=$ $=$	$- 8$ $-8$

$x$	$:$	$x,1$
$x+2$	$:$	$(x-2)$ , $1$
$x^2-4$	$:$	$(x-2)$ , $(x+2)$

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

$-3 – 2$	$=$	$-5$
$-3 xx -2$	$=$	$6$

$-3 + 2$	$=$	$-1$
$-3 xx 2$	$=$	$-6$

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