Factorise Trinomials (Quadratics) w Coefficient more than 1 (2)

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Question 1 of 4

1. Question
Factorise.

$5 x^{2} + 22 x + 8$ $5x^2+22x+8$
- 1. $(5 x + 15) (x + 7)$ $(5x+15)(x+7)$
- 2. $(5 x + 2) (x + 11)$ $(5x+2)(x+11)$
- 3. $(5 x + 1) (x + 8)$ $(5x+1)(x+8)$
- 4. $(5 x + 2) (x + 4)$ $(5x+2)(x+4)$
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When factorising trinomials, use the Cross Method.

Use the cross method to factorise $5 x^{2} + 22 x + 8$ $5x^2+22x+8$

Start by drawing a cross.

Now, find two values that will multiply into $5 x^{2}$ $5x^2$ and write them on the left side of the cross.

$5 x$ $5x$ and $x$ $x$ fits this description.

Next, find two numbers that will multiply into $8$ $8$ and, when cross-multiplied to the values to the left side, will add up to $22 x$ $22x$ .

Product Sum when Cross-Multiplied

$2$ $2$ and $4$ $4$ $8$ $8$ $(5 x \times 4) + (x \times 2) = 22 x$ $(5xtimes4)+(xtimes2)=22x$

$8$ $8$ and $1$ $1$ $8$ $8$ $(5 x \times 1) + (x \times 8) = 13 x$ $(5xtimes1)+(xtimes8)=13x$

$2$ $2$ and $4$ $4$ fits this description.

Now, write $2$ $2$ and $4$ $4$ 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 $(5 x + 2) (x + 4)$ $(5x+2)(x+4)$ .

$(5 x + 2) (x + 4)$ $(5x+2)(x+4)$
Question 2 of 4

2. Question
Factorise.

$3 x^{2} - 26 x + 35$ $3x^2-26x+35$
- 1. $(3 x - 5) (x - 7)$ $(3x-5)(x-7)$
- 2. $(3 x - 2) (2 x - 13)$ $(3x-2)(2x-13)$
- 3. $(3 x - 7) (x - 5)$ $(3x-7)(x-5)$
- 4. $(2 x - 3) (x + 7)$ $(2x-3)(x+7)$
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When factorising trinomials, use the Cross Method.

Use the cross method to factorise $3x^2-26x+35$

Start by drawing a cross.

Now, find two values that will multiply into $3x^2$ and write them on the left side of the cross.

$3x$ and $x$ fits this description.

Next, find two numbers that will multiply into $35$ and, when cross-multiplied to the values to the left side, will add up to $-26x$ .

Product Sum when Cross-Multiplied

$-7$ and $-5$ $35$ $[3xtimes(-5)]+[xtimes(-7)]=-22x$

$-5$ and $-7$ $35$ $[3xtimes(-7)]+[xtimes(-5)]=-26x$

$-5$ and $-7$ fits this description.

Now, write $-5$ and $-7$ 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 $(3x-5)(x-7)$ .

$(3x-5)(x-7)$
Question 3 of 4

3. Question
Factorise.

$6x^2-17x-3$
- 1. $(6x+1)(x-3)$
- 2. $(3x+1)(3x-3)$
- 3. $(6x+3)(x-1)$
- 4. $(2x+1)(3x-3)$
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When factorising trinomials, use the Cross Method.

Use the cross method to factorise $6x^2-17x-3$

Start by drawing a cross.

Now, find two values that will multiply into $6x^2$ and write them on the left side of the cross.

$6x$ and $x$ 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 $-17x$ .

Product Sum when Cross-Multiplied

$1$ and $-3$ $-3$ $[6xtimes(-3)]+(xtimes1)=-17x$

$3$ and $-1$ $-3$ $(6xtimes1)+[xtimes(-6)]=-5x$

$1$ 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 $(6x+1)(x-3)$ .

$(6x+1)(x-3)$

	Product	Sum when Cross-Multiplied
$2$ $2$ and $4$ $4$	$8$ $8$	$(5 x \times 4) + (x \times 2) = 22 x$ $(5xtimes4)+(xtimes2)=22x$
$8$ $8$ and $1$ $1$	$8$ $8$	$(5 x \times 1) + (x \times 8) = 13 x$ $(5xtimes1)+(xtimes8)=13x$

	Product	Sum when Cross-Multiplied
$-7$ and $-5$	$35$	$[3xtimes(-5)]+[xtimes(-7)]=-22x$
$-5$ and $-7$	$35$	$[3xtimes(-7)]+[xtimes(-5)]=-26x$

	Product	Sum when Cross-Multiplied
$1$ and $-3$	$-3$	$[6xtimes(-3)]+(xtimes1)=-17x$
$3$ and $-1$	$-3$	$(6xtimes1)+[xtimes(-6)]=-5x$

Question 4 of 4

Factorise.

$5u^2+19u+12$

1. $(5u+3)(u+4)$
2. $(5u+1)(u+12)$
3. $(5u+2)(u+6)$
4. $(5u+4)(u+3)$

	Product	Sum when Cross-Multiplied
$2$ and $6$	$12$	$(5utimes6)+(utimes2)=32u$
$3$ and $4$	$12$	$(5utimes4)+(utimes3)=23u$
$4$ and $3$	$12$	$(5utimes3)+(utimes4)=19u$
$1$ and $12$	$12$	$(5utimes12)+(utimes1)=61u$

Factorise Trinomials (Quadratics) w Coefficient more than 1 (2)

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