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

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

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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

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

$6x^2-11x+3$

Start by drawing a cross.

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

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

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

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

$-11x$ .

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

$3x$ and

$2x$ fits the left side and

$-1$ and

$-3$ 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

$HCF$ before the brackets.

Therefore, the factorised expression is

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

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

Question 3 of 4

Factorise.

$12y^2-24y-63$

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

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

Question 4 of 4

Factorise.

$15-u-2u^2$

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

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

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

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Quizzes

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