Factor Quadratics with Leading Coefficient more than 1 (1)

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Question 1 of 4

1. Question
Factorise.

$3 m^{2} + 24 m + 36$ $3m^2+24m+36$
- 1. $3 (m - 4) (m + 6)$ $3(m-4)(m+6)$
- 2. $3 (m + 4) (m - 8)$ $3(m+4)(m-8)$
- 3. $4 (m - 3) (m + 10)$ $4(m-3)(m+10)$
- 4. $3 (m + 2) (m + 6)$ $3(m+2)(m+6)$
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When factorising trinomials, use the Cross Method.

First, find the Highest Common Factor (HCF) of the three terms.

Start by listing down their factors.

Factors of $3 m^{2}$ $3m^2$ : $3$ $3$ $\times m \times m$ $timesmtimesm$

Factors of $24 m$ $24m$ : $3$ $3$ $\times 8 \times m$ $times8timesm$

Factors of $36$ $36$ : $3$ $3$ $\times 12$ $times12$

All the terms have $3$ $3$ as their factor, so it is the HCF.

Next, factorise by placing $3$ $3$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by $3$ $3$ , then simplify.

$3 [(3 m^{2} \div 3) + (24 m \div 3) + (36 \div 3)]$ $3[(3m^2div3)+(24mdiv3)+(36div3)]$

$=$ $=$ $3 (m^{2} + 8 m + 12)$ $3(m^2+8m+12)$

Now, use the cross method to factorise $m^{2} + 8 m + 12$ $m^2+8m+12$

Start by drawing a cross.

Then, find two numbers that will multiply into $12$ $12$ and add up to $8$ $8$

Product Sum

$3$ $3$ and $4$ $4$ $12$ $12$ $7$ $7$

$2$ $2$ and $6$ $6$ $12$ $12$ $8$ $8$

$2$ $2$ and $6$ $6$ fits this description.

Write $2$ $2$ and $6$ $6$ on the right side of the cross.

Now, find two values that will multiply into $m^{2}$ $m^2$ and write them on the left side of the cross.

$m$ $m$ and $m$ $m$ fits this description.

Finally, group the values in a row with a bracket and combine the brackets.

Remember to add the $H C F$ $HCF$ before the brackets.

Therefore, the factorised expression is $3 (m + 2) (m + 6)$ $3(m+2)(m+6)$ .

$3 (m + 2) (m + 6)$ $3(m+2)(m+6)$
Question 2 of 4

2. Question
Factorise.

$5 b^{2} - 30 b - 135$ $5b^2-30b-135$
- 1. $5 (b + 27) (b - 1)$ $5(b+27)(b-1)$
- 2. $5 (b + 1) (b - 27)$ $5(b+1)(b-27)$
- 3. $5 (b + 9) (b - 3)$ $5(b+9)(b-3)$
- 4. $5 (b + 3) (b - 9)$ $5(b+3)(b-9)$
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When factorising trinomials, use the Cross Method.

First, find the Highest Common Factor (HCF) of the three terms.

Start by listing down their factors.

Factors of $5 b^{2}$ $5b^2$ : $5$ $5$ $\times b \times b$ $timesbtimesb$

Factors of $30 b$ $30b$ : $5$ $5$ $\times 6 \times b$ $times6timesb$

Factors of $135$ $135$ : $5$ $5$ $\times 27$ $times27$

All the terms have $5$ $5$ as their factor, so it is the HCF.

Next, factorise by placing $5$ $5$ outside a bracket.

Also, place the given polynomial inside the bracket with each term divided by $5$ $5$ , then simplify.

$5 [(5 b^{2} \div 5) - (30 b \div 5) - (135 \div 5)]$ $5[(5b^2div5)-(30bdiv5)-(135div5)]$

$=$ $=$ $5 (b^{2} - 6 b - 27)$ $5(b^2-6b-27)$

Now, use the cross method to factorise $b^{2} - 6 b - 27$ $b^2-6b-27$

Start by drawing a cross.

Then, find two numbers that will multiply into $- 27$ $-27$ and add up to $- 6$ $-6$

Product Sum

$1$ $1$ and $- 27$ $-27$ $- 27$ $-27$ $- 26$ $-26$

$3$ $3$ and $- 9$ $-9$ $- 27$ $-27$ $- 6$ $-6$

$3$ $3$ and $- 9$ $-9$ fits this description.

Write $3$ $3$ and $- 9$ $-9$ on the right side of the cross.

Now, find two values that will multiply into $b^{2}$ $b^2$ and write them on the left side of the cross.

$b$ $b$ and $b$ $b$ fits this description.

Finally, group the values in a row with a bracket and combine the brackets.

Remember to add the $H C F$ $HCF$ before the brackets.

Therefore, the factorised expression is $5 (b + 3) (b - 9)$ $5(b+3)(b-9)$ .

$5 (b + 3) (b - 9)$ $5(b+3)(b-9)$
Question 3 of 4

3. Question
Factorise.

$4 x^{2} - 32 x + 60$ $4x^2-32x+60$
- 1. $4 (x - 7) (x - 8)$ $4(x-7)(x-8)$
- 2. $4 (x + 3) (x - 5)$ $4(x+3)(x-5)$
- 3. $4 (x - 1) (x - 15)$ $4(x-1)(x-15)$
- 4. $4 (x - 3) (x - 5)$ $4(x-3)(x-5)$
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When factorising trinomials, use the Cross Method.

First, find the Highest Common Factor (HCF) of the three terms.

Start by listing down their factors.

Factors of $4 x^{2}$ $4x^2$ : $4$ $4$ $\times x \times x$ $timesxtimesx$

Factors of $32 x$ $32x$ : $4$ $4$ $\times 8 \times x$ $times8timesx$

Factors of $60$ $60$ : $4$ $4$ $\times 15$ $times15$

All the terms have $4$ $4$ as their factor, so it is the HCF.

Next, factorise by placing $4$ $4$ outside a bracket.

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

$4 [(4 x^{2} \div 4) - (32 x \div 4) + (60 \div 4)]$ $4[(4x^2div4)-(32xdiv4)+(60div4)]$

$=$ $=$ $4 (x^{2} - 8 x + 15)$ $4(x^2-8x+15)$

Now, use the cross method to factorise $x^{2} - 8 x + 15$ $x^2-8x+15$

Start by drawing a cross.

Then, find two numbers that will multiply into $15$ $15$ and add up to $- 8$ $-8$

Product Sum

$1$ $1$ and $- 15$ $-15$ $- 15$ $-15$ $- 14$ $-14$

$- 3$ $-3$ and $- 5$ $-5$ $15$ $15$ $- 6$ $-6$

$- 3$ $-3$ and $- 5$ $-5$ fits this description.

Write $- 3$ $-3$ and $- 5$ $-5$ on the right side of the cross.

Now, find two values that will multiply into $x^{2}$ $x^2$ and write them on the left side of the cross.

$x$ $x$ and $x$ $x$ fits this description.

Finally, group the values in a row with a bracket and combine the brackets.

Remember to add the $H C F$ $HCF$ before the brackets.

Therefore, the factorised expression is $4 (x - 3) (x - 5)$ $4(x-3)(x-5)$ .

$4 (x - 3) (x - 5)$ $4(x-3)(x-5)$
Question 4 of 4

4. Question
Factorise.

$2 m^{2} + 7 m + 3$ $2m^2+7m+3$
- 1. $(2 m + 10) (m - 3)$ $(2m+10)(m-3)$
- 2. $(2 m + 1) (m + 3)$ $(2m+1)(m+3)$
- 3. $(2 m + 3) (m + 3)$ $(2m+3)(m+3)$
- 4. $(2 m + 4) (2 m + 3)$ $(2m+4)(2m+3)$
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When factorising trinomials, use the Cross Method.

Use the cross method to factorise $2 m^{2} + 7 m + 3$ $2m^2+7m+3$

Start by drawing a cross.

Now, find two values that will multiply into $2 m^{2}$ $2m^2$ and write them on the left side of the cross.

$2 m$ $2m$ and $m$ $m$ fits this description.

Next, find two numbers that will multiply into $3$ $3$ and, when cross-multiplied to the values to the left side, will add up to $7 m$ $7m$ .

Product Sum when Cross-Multiplied

$3$ $3$ and $1$ $1$ $3$ $3$ $(2 m \times 1) + (m \times 3) = 4 m$ $(2mtimes1)+(mtimes3)=4m$

$1$ $1$ and $3$ $3$ $3$ $3$ $(2 m \times 3) + (m \times 1) = 7 m$ $(2mtimes3)+(mtimes1)=7m$

$1$ $1$ and $3$ $3$ fits this description.

Now, write $1$ $1$ and $3$ $3$ on the right side of the cross.

Finally, group the values in a row with a bracket and combine the brackets.

Therefore, the factorised expression is $(2 m + 1) (m + 3)$ $(2m+1)(m+3)$ .

$(2 m + 1) (m + 3)$ $(2m+1)(m+3)$

Factor Quadratics with Leading Coefficient more than 1 (1)

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Quizzes

		$3 [(3 m^{2} \div 3) + (24 m \div 3) + (36 \div 3)]$ $3[(3m^2div3)+(24mdiv3)+(36div3)]$
	$=$ $=$	$3 (m^{2} + 8 m + 12)$ $3(m^2+8m+12)$

	Product	Sum
$3$ $3$ and $4$ $4$	$12$ $12$	$7$ $7$
$2$ $2$ and $6$ $6$	$12$ $12$	$8$ $8$

		$5 [(5 b^{2} \div 5) - (30 b \div 5) - (135 \div 5)]$ $5[(5b^2div5)-(30bdiv5)-(135div5)]$
	$=$ $=$	$5 (b^{2} - 6 b - 27)$ $5(b^2-6b-27)$

	Product	Sum
$1$ $1$ and $- 27$ $-27$	$- 27$ $-27$	$- 26$ $-26$
$3$ $3$ and $- 9$ $-9$	$- 27$ $-27$	$- 6$ $-6$

		$4 [(4 x^{2} \div 4) - (32 x \div 4) + (60 \div 4)]$ $4[(4x^2div4)-(32xdiv4)+(60div4)]$
	$=$ $=$	$4 (x^{2} - 8 x + 15)$ $4(x^2-8x+15)$

	Product	Sum
$1$ $1$ and $- 15$ $-15$	$- 15$ $-15$	$- 14$ $-14$
$- 3$ $-3$ and $- 5$ $-5$	$15$ $15$	$- 6$ $-6$

	Product	Sum when Cross-Multiplied
$3$ $3$ and $1$ $1$	$3$ $3$	$(2 m \times 1) + (m \times 3) = 4 m$ $(2mtimes1)+(mtimes3)=4m$
$1$ $1$ and $3$ $3$	$3$ $3$	$(2 m \times 3) + (m \times 1) = 7 m$ $(2mtimes3)+(mtimes1)=7m$