Factorise Trinomials (Quadratics) w Coefficient more than 1 (3)

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Question 1 of 4

Factorise.

10 x^{2} + 29 x - 72

$10x^2+29x-72$

1. $(5 x - 2) (2 x + 36)$ $(5x-2)(2x+36)$
2. $(5 x - 8) (2 x + 9)$ $(5x-8)(2x+9)$
3. $(5 x - 12) (2 x + 6)$ $(5x-12)(2x+6)$
4. $(5 x + 3) (2 x - 24)$ $(5x+3)(2x-24)$

	Product	Sum when Cross-Multiplied
$- 9$ $-9$ and $8$ $8$	$- 72$ $-72$	$(5 x \times 8) + [2 x \times (- 9)] = 22 x$ $(5xtimes8)+[2xtimes(-9)]=22x$
$- 8$ $-8$ and $9$ $9$	$- 72$ $-72$	$(5 x \times 9) + [2 x \times (- 8)] = 29 x$ $(5xtimes9)+[2xtimes(-8)]=29x$

Question 2 of 4

Factorise.

18 x^{2} - 33 x + 9

$18x^2-33x+9$

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

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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

18 x^{2}

$18x^2$ : $3$ $3$

\times 6 \times x \times x

$times6timesxtimesx$

Factors of

- 33 x

$-33x$ : $3$ $3$

\times - 11 \times x

$times-11timesx$

Factors of

9

$9$ : $3$ $3$

\times 3

$times3$

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.

		$3 [(18 x^{2} \div 3) - (33 x \div 3) + (9 \div 3)]$ $3[(18x^2div3)-(33xdiv3)+(9div3)]$
	$=$ $=$	$3 (6 x^{2} - 11 x - 3)$ $3(6x^2-11x-3)$

Now, use the cross method to factorise

6 x^{2} - 11 x + 3

$6x^2-11x+3$

Start by drawing a cross.

For the left side, find two values that will multiply into

6 x^{2}

$6x^2$ and write them on the left side of the cross.

While for the right side, find two numbers that will multiply into

3

$3$ and, when cross-multiplied to the values to the left side, will add up to

- 11 x

$-11x$ .

Left Side	Product	Right Side	Product	Sum when Cross-Multiplied
$6 x$ $6x$ and $x$ $x$	$6 x^{2}$ $6x^2$	$- 3$ $-3$ and $- 1$ $-1$	$3$ $3$	$(6 x \times - 1) + (x \times - 3) = - 3 x$ $(6xtimes-1)+(xtimes-3)=-3x$
$3 x$ $3x$ and $2 x$ $2x$	$6 x^{2}$ $6x^2$	$- 3$ $-3$ and $- 1$ $-1$	$3$ $3$	$(3 x \times - 1) + (2 x \times - 3) = - 9 x$ $(3xtimes-1)+(2xtimes-3)=-9x$
$3 x$ $3x$ and $2 x$ $2x$	$6 x^{2}$ $6x^2$	$- 1$ $-1$ and $- 3$ $-3$	$3$ $3$	$(3 x \times - 3) + (2 x \times - 1) = - 11 x$ $(3xtimes-3)+(2xtimes-1)=-11x$

3 x

$3x$ and

2 x

$2x$ fits the left side and

- 1

$-1$ and

- 3

$-3$ fits the right side.

Now, write the chosen values on the sides of the cross.

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 (3 x - 1) (2 x - 3)

$3(3x-1)(2x-3)$ .

3 (3 x - 1) (2 x - 3)

$3(3x-1)(2x-3)$

Question 3 of 4

Factorise.

12 y^{2} - 24 y - 63

$12y^2-24y-63$

1. $3 (4 y - 3) (y + 7)$ $3(4y-3)(y+7)$
2. $3 (2 y - 3) (2 y + 7)$ $3(2y-3)(2y+7)$
3. $3 (2 y + 3) (2 y - 7)$ $3(2y+3)(2y-7)$
4. $3 (4 y + 3) (y - 7)$ $3(4y+3)(y-7)$

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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

12 y^{2}

$12y^2$ : $3$ $3$

\times 4 \times y \times y

$times4timesytimesy$

Factors of

24 y

$24y$ : $3$ $3$

\times 8 \times y

$times8timesy$

Factors of

63

$63$ : $3$ $3$

\times 21

$times21$

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.

		$3 [(12 y^{2} \div 3) - (24 y \div 3) - (63 \div 3)]$ $3[(12y^2div3)-(24ydiv3)-(63div3)]$
	$=$ $=$	$3 (4 y^{2} - 8 y - 21)$ $3(4y^2-8y-21)$

Now, use the cross method to factorise

4 y^{2} - 8 y - 21

$4y^2-8y-21$

Start by drawing a cross.

For the left side, find two values that will multiply into

4 y^{2}

$4y^2$ and write them on the left side of the cross.

While for the right side, find two numbers that will multiply into

- 21

$-21$ and, when cross-multiplied to the values to the left side, will add up to

- 8 y

$-8y$ .

Left Side	Product	Right Side	Product	Sum when Cross-Multiplied
$4 y$ $4y$ and $y$ $y$	$4 y^{2}$ $4y^2$	$3$ $3$ and $- 7$ $-7$	$- 21$ $-21$	$(4 y \times - 7) + (3 \times y) = - 25 y$ $(4ytimes-7)+(3timesy)=-25y$
$2 y$ $2y$ and $2 y$ $2y$	$4 y^{2}$ $4y^2$	$3$ $3$ and $- 7$ $-7$	$- 21$ $-21$	$(2 y \times - 7) + (2 y \times 3) = - 8 y$ $(2ytimes-7)+(2ytimes3)=-8y$

2 y

$2y$ and

2 y

$2y$ fits the left side and

3

$3$ and

- 7

$-7$ fits the right side.

Now, write the chosen values on the sides of the cross.

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 (2 y + 3) (2 y - 7)

$3(2y+3)(2y-7)$ .

3 (2 y + 3) (2 y - 7)

$3(2y+3)(2y-7)$

Question 4 of 4

Factorise.

15 - u - 2 u^{2}

$15-u-2u^2$

1. $(3 + u) (5 - 2 u)$ $(3+u)(5-2u)$
2. $(5 - u) (3 + 2 u)$ $(5-u)(3+2u)$
3. $(3 - u) (5 + 2 u)$ $(3-u)(5+2u)$
4. $(5 + u) (3 - 2 u)$ $(5+u)(3-2u)$

Left Side	Product	Right Side	Product	Sum when Cross-Multiplied
$3$ $3$ and $5$ $5$	$15$ $15$	$2 u$ $2u$ and $- u$ $-u$	$- 2 u^{2}$ $-2u^2$	$(3 \times - u) + (5 \times 2 u) = 7 u$ $(3times-u)+(5times2u)=7u$
$3$ $3$ and $5$ $5$	$15$ $15$	$u$ $u$ and $- 2 u$ $-2u$	$- 2 u^{2}$ $-2u^2$	$(3 \times - 2 u) + (5 \times u) = - u$ $(3times-2u)+(5timesu)=-u$

Factorise Trinomials (Quadratics) w Coefficient more than 1 (3)

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