Factor Quadratics with Leading Coefficient more than 1 (1)

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Question 1 of 4

1. Question
Factorise.

$3m^2+24m+36$
- 1. $3(m+2)(m+6)$
- 2. $4(m-3)(m+10)$
- 3. $3(m-4)(m+6)$
- 4. $3(m+4)(m-8)$
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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 $3m^2$ : $3$ $timesmtimesm$

Factors of $24m$ : $3$ $times8timesm$

Factors of $36$ : $3$ $times12$

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

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

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

$3[(3m^2div3)+(24mdiv3)+(36div3)]$

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

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

Start by drawing a cross.

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

Product Sum

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

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

$2$ and $6$ fits this description.

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

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

$m$ and $m$ fits this description.

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

Remember to add the $HCF$ before the brackets.

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

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

2. Question
Factorise.

$5b^2-30b-135$
- 1. $5(b+3)(b-9)$
- 2. $5(b+27)(b-1)$
- 3. $5(b+1)(b-27)$
- 4. $5(b+9)(b-3)$
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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 $5b^2$ : $5$ $timesbtimesb$

Factors of $30b$ : $5$ $times6timesb$

Factors of $135$ : $5$ $times27$

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

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

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

$5[(5b^2div5)-(30bdiv5)-(135div5)]$

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

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

Start by drawing a cross.

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

Product Sum

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

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

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

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

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

$b$ and $b$ fits this description.

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

Remember to add the $HCF$ before the brackets.

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

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

3. Question
Factorise.

$4x^2-32x+60$
- 1. $4(x+3)(x-5)$
- 2. $4(x-7)(x-8)$
- 3. $4(x-1)(x-15)$
- 4. $4(x-3)(x-5)$
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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 $4x^2$ : $4$ $timesxtimesx$

Factors of $32x$ : $4$ $times8timesx$

Factors of $60$ : $4$ $times15$

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

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

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

$4[(4x^2div4)-(32xdiv4)+(60div4)]$

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

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

Start by drawing a cross.

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

Product Sum

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

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

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

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

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

$x$ and $x$ fits this description.

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

Remember to add the $HCF$ before the brackets.

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

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

4. Question
Factorise.

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

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

Start by drawing a cross.

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

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

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

Product Sum when Cross-Multiplied

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

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

$1$ and $3$ fits this description.

Now, write $1$ and $3$ 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 $(2m+1)(m+3)$ .

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

Factor Quadratics with Leading Coefficient more than 1 (1)

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Quizzes

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

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

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

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

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