Factorise a Polynomial with Integers

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Question 1 of 6

1. Question
Factor.

$- 18 x + 9$ $-18x+9$
- 1. $- 4 (4 x - 2)$ $-4(4x-2)$
- 2. $9 (- x + 9)$ $9(-x+9)$
- 3. $- 3 (4 x + 1)$ $-3(4x+1)$
- 4. $- 9 (2 x - 1)$ $-9(2x-1)$
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A Greatest Common Factor is the factor of two terms with the highest value.

First, find the Greatest Common Factor (GCF) of the two terms.

Factors of $18 x$ $18x$ : $1$ $1$ $\times 2 \times$ $times2times$ $9$ $9$ $\times x$ $times x$

Factors of $9$ $9$ : $1$ $1$ $\times$ $times$ $9$ $9$

Collect the common factors and multiply them all to get the GCF.

GCF $=$ $=$ $1$ $1$ $\times$ $times$ $9$ $9$

$=$ $=$ $9$ $9$

Since the first term is negative, we can make the GCF negative as well to make the factorising easier.

Therefore, the GCF that we will use is $- 9$ $-9$ .

Finally, factor by placing $- 9$ $-9$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by $- 9$ $-9$ , then simplify.

$- 9 [(- 18 x \div (- 9)) + (9 \div (- 9))]$ $-9[(-18xdiv(-9))+(9div(-9))]$

$=$ $=$ $- 9 [2 x + (- 1)]$ $-9[2x+(-1)]$

$=$ $=$ $- 9 (2 x - 1)$ $-9(2x-1)$

$- 9 (2 x - 1)$ $-9(2x-1)$
Question 2 of 6

2. Question
Factor.

$- 2 m - 6$ $-2m-6$
- 1. $- 2 (m + 3)$ $-2(m+3)$
- 2. $- 2 (2 m - 6)$ $-2(2m-6)$
- 3. $- 2 (m + 2)$ $-2(m+2)$
- 4. $m (m - 6)$ $m(m-6)$
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A Greatest Common Factor is the factor of two terms with the highest value.

First, find the Greatest Common Factor (GCF) of the two terms.

Factors of $2 m$ $2m$ : $2$ $2$ $\times m$ $times m$

Factors of $6$ $6$ : $2$ $2$ $\times 3$ $times3$

Since the first term is negative, we can make the GCF negative as well to make the factorising easier.

Therefore, the GCF that we will use is $- 2$ $-2$ .

Finally, factor by placing $- 2$ $-2$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by $- 2$ $-2$ , then simplify.

$- 2 [(- 2 m \div (- 2)) - (6 \div (- 2))]$ $-2[(-2mdiv(-2))-(6div(-2))]$

$=$ $=$ $- 2 [m - (- 3)]$ $-2[m-(-3)]$

$=$ $=$ $- 2 (m + 3)$ $-2(m+3)$

$- 2 (m + 3)$ $-2(m+3)$

Question 3 of 6

Factor.

- 6 a b^{2} + 30 a b

$-6ab^2+30ab$

1. $- 8 b (a b - 6)$ $-8b(ab-6)$
2. $- 5 a b (2 b - 6)$ $-5ab(2b-6)$
3. $- 4 a b (2 b - 7)$ $-4ab(2b-7)$
4. $- 6 a b (b - 5)$ $-6ab(b-5)$

Question 4 of 6

4. Question
Factor.

$- 20 m - 10 m^{2}$ $-20m-10m^2$
- 1. $- 4 m (5 - 3 m)$ $-4m(5-3m)$
- 2. $- 10 m (2 + m)$ $-10m(2+m)$
- 3. $- 5 m (5 - 2 m)$ $-5m(5-2m)$
- 4. $- 10 (2 m - m)$ $-10(2m-m)$
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A Greatest Common Factor is the factor of two terms with the highest value.

First, find the Greatest Common Factor (GCF) of the two terms.

Factors of $20 m$ $20m$ : $2 \times$ $2times$ $10$ $10$ $\times$ $times$ $m$ $m$

Factors of $10 m^{2}$ $10m^2$ : $1 \times$ $1times$ $10$ $10$ $\times$ $times$ $m$ $m$ $\times m$ $times m$

Collect the common factors and multiply them all to get the GCF.

GCF $=$ $=$ $10$ $10$ $\times$ $times$ $m$ $m$

$=$ $=$ $10 m$ $10m$

Since the first term is negative, we can make the GCF negative as well to make the factorising easier.

Therefore, the GCF that we will use is $- 10 m$ $-10m$ .

Finally, factor by placing $- 10 m$ $-10m$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by $- 10 m$ $-10m$ , then simplify.

$- 10 m [(- 20 m \div (- 10 m)) - (10 m^{2} \div (- 10 m))]$ $-10m[(-20mdiv(-10m))-(10m^2div(-10m))]$

$=$ $-10m(2+m)$

$-10m(2+m)$
Question 5 of 6

5. Question
Factor.

$-8x^2+20x$
- 1. $-4x(2x-5)$
- 2. $-5x(2x+4)$
- 3. $-4(3x-5x)$
- 4. $-6x(2x+2)$
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A Greatest Common Factor is the factor of two terms with the highest value.

First, find the Greatest Common Factor (GCF) of the two terms.

Factors of $8x^2$ : $2times$ $4$ $times$ $x$ $times x$

Factors of $20x$ : $4$ $times5times$ $x$

Collect the common factors and multiply them all to get the GCF.

GCF $=$ $4$ $times$ $x$

$=$ $4x$

Since the first term is negative, we can make the GCF negative as well to make the factorising easier.

Therefore, the GCF that we will use is $-4x$ .

Finally, factor by placing $-4x$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by $-4x$ , then simplify.

$-4x[(-8x^2div(-4x))+(20xdiv(-4x))]$

$=$ $-4x[2x+(-5)]$

$=$ $-4x(2x-5)$

$-4x(2x-5)$
Question 6 of 6

6. Question
Factor.

$-26b^2c-39bc^2$
- 1. $13bc^2(2b+3)$
- 2. $-13b^2c(2+3c)$
- 3. $13b(3b+2c)$
- 4. $-13bc(2b+3c)$
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A Greatest Common Factor is the factor of two terms with the highest value.

First, find the Greatest Common Factor (GCF) of the two terms.

Start by listing down their factors.

Factors of $26b^2c$ : $2times$ $13$ $times$ $b$ $times b times$ $c$

Factors of $39bc^2$ : $3times$ $13$ $times$ $b$ $times$ $c$ $times c$

Collect the common factors and multiply them all to get the GCF.

GCF $=$ $13$ $times$ $b$ $times$ $c$

$=$ $13bc$

Since the first term is negative, we can make the GCF negative as well to make the factorising easier.

Therefore, the GCF that we will use is $-13bc$ .

Finally, factor by placing $-13bc$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by $-13bc$ , then simplify.

$-13bc[(-26b^2cdiv-13bc)+(-39bc^2div-13bc)]$

$=$ $-13bc[2b+(3c)]$

$=$ $-13bc(2b+3c)$

$-13bc(2b+3c)$

Factorise a Polynomial with Integers

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Quiz summary

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Quizzes

GCF	$=$ $=$	$1$ $1$ $\times$ $times$ $9$ $9$
	$=$ $=$	$9$ $9$

		$- 9 [(- 18 x \div (- 9)) + (9 \div (- 9))]$ $-9[(-18xdiv(-9))+(9div(-9))]$
	$=$ $=$	$- 9 [2 x + (- 1)]$ $-9[2x+(-1)]$
	$=$ $=$	$- 9 (2 x - 1)$ $-9(2x-1)$

		$- 2 [(- 2 m \div (- 2)) - (6 \div (- 2))]$ $-2[(-2mdiv(-2))-(6div(-2))]$
	$=$ $=$	$- 2 [m - (- 3)]$ $-2[m-(-3)]$
	$=$ $=$	$- 2 (m + 3)$ $-2(m+3)$

GCF	$=$ $=$	$6$ $6$ $\times$ $times$ $a$ $a$ $\times$ $times$ $b$ $b$
	$=$ $=$	$6 a b$ $6ab$

		$- 6 a b [(- 6 a b^{2} \div (- 6 a b)) + (30 a b \div (- 6 a b))]$ $-6ab[(-6ab^2div(-6ab))+(30abdiv(-6ab))]$
	$=$ $=$	$- 6 a b [b + (- 5)]$ $-6ab[b+(-5)]$
	$=$ $=$	$- 6 a b (b - 5)$ $-6ab(b-5)$

GCF	$=$ $=$	$10$ $10$ $\times$ $times$ $m$ $m$
	$=$ $=$	$10 m$ $10m$

		$- 10 m [(- 20 m \div (- 10 m)) - (10 m^{2} \div (- 10 m))]$ $-10m[(-20mdiv(-10m))-(10m^2div(-10m))]$
	$=$	$-10m(2+m)$

		$-4x[(-8x^2div(-4x))+(20xdiv(-4x))]$
	$=$	$-4x[2x+(-5)]$
	$=$	$-4x(2x-5)$

		$-13bc[(-26b^2cdiv-13bc)+(-39bc^2div-13bc)]$
	$=$	$-13bc[2b+(3c)]$
	$=$	$-13bc(2b+3c)$