Solve Quadratics by Factoring

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Question 1 of 4

1. Question
Solve for $x$ $x$ :

$x^{2} - 6 x + 5 = 0$ $x^2-6x+5=0$
- 1. $1$ $1$ and $5$ $5$
- 2. $- 1$ $-1$ and $- 5$ $-5$
- 3. $- 1$ $-1$ and $5$ $5$
- 4. $1$ $1$ and $- 5$ $-5$
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The cross method is a factorisation method used for quadratics.

Since the equation is in standard form $($ $($ $a$ $a$ $x^{2} +$ $x^2+$ $b$ $b$ $x +$ $x+$ $c$ $c$ $= 0)$ $=0)$ we can factorise using the cross method.

$x^{2}$ $x^2$ $- 6$ $-6$ $x +$ $x+$ $5$ $5$ $= 0$ $=0$

To factorise, we need to find two numbers that add to $- 6$ $-6$ and multiply to $5$ $5$

$- 5$ $-5$ and $- 1$ $-1$ fit both conditions

$- 5 + (- 1)$ $-5 + (-1)$ $=$ $=$ $- 6$ $-6$

$- 5 \times - 1$ $-5 xx -1$ $=$ $=$ $5$ $5$

Read across to get the factors.

$(x - 5) (x - 1) = 0$ $(x-5)(x-1)=0$

Solve for each value of $x$ $x$ .

$x - 5$ $x-5$ $=$ $=$ $0$ $0$

$x - 5$ $x-5$ $+ 5$ $+5$ $=$ $=$ $0$ $0$ $+ 5$ $+5$

$x$ $x$ $=$ $=$ $5$ $5$

$x - 1$ $x-1$ $=$ $=$ $0$ $0$

$x - 1$ $x-1$ $+ 1$ $+1$ $=$ $=$ $0$ $0$ $+ 1$ $+1$

$x$ $x$ $=$ $=$ $1$ $1$

$x = 1, 5$ $x=1, 5$
Question 2 of 4

2. Question
Solve for $x$ $x$ :

$2 x^{2} - 3 x + 1 = 0$ $2x^2-3x+1=0$
- 1. $\frac{1}{2}$ $1/2$ and $1$ $1$
- 2. $- \frac{1}{2}$ $-1/2$ and $- 1$ $-1$
- 3. $- \frac{1}{2}$ $-1/2$ and $1$ $1$
- 4. $\frac{1}{2}$ $1/2$ and $- 1$ $-1$
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The cross method is a factorisation method used for quadratics.

Since the equation is in standard form $($ $($ $a$ $a$ $x^{2} +$ $x^2+$ $b$ $b$ $x +$ $x+$ $c$ $c$ $= 0)$ $=0)$ we can factorise using the cross method.

$2$ $2$ $x^{2}$ $x^2$ $- 3$ $-3$ $x$ $x$ $+ 1$ $+1$ $= 0$ $=0$

Find two numbers that multiply to $2$ $2$ on the left side and multiply to $1$ $1$ on the right side. Also, when cross multiplied they should add to the middle term ( $- 3 x$ $-3x$ ).

$2$ $2$ and $1$ $1$ fit the condition for the left side and two $- 1$ $-1$ ’s fit the condition for the right side

$2 \times 1$ $2 xx 1$ $=$ $=$ $2$ $2$

$- 1 \times - 1$ $-1 xx -1$ $=$ $=$ $1$ $1$

$(2 x \times - 1) + (x \times - 1)$ $(2x xx -1) + (x xx -1)$ $=$ $=$ $- 3 x$ $-3x$

Read across to get the factors.

$(2 x - 1) (x - 1) = 0$ $(2x-1)(x-1)=0$

Solve for each value of $x$ $x$ .

$2 x - 1$ $2x-1$ $=$ $=$ $0$ $0$

$2 x - 1$ $2x-1$ $+ 1$ $+1$ $=$ $=$ $0$ $0$ $+ 1$ $+1$

$2 x$ $2x$ $=$ $=$ $1$ $1$

$x$ $x$ $=$ $=$ $\frac{1}{2}$ $1/2$

$x - 1$ $x-1$ $=$ $=$ $0$ $0$

$x - 1$ $x-1$ $+ 1$ $+1$ $=$ $=$ $0$ $0$ $+ 1$ $+1$

$x$ $x$ $=$ $=$ $1$ $1$

$x = \frac{1}{2}, 1$ $x=1/2, 1$
Question 3 of 4

3. Question
Solve for $x$ $x$ :

$3 x^{2} + 2 x - 21 = 0$ $3x^2+2x-21=0$
- 1. $2 \frac{1}{3}$ $2 1/3$ and $3$ $3$
- 2. $- 2 \frac{1}{3}$ $-2 1/3$ and $3$
- 3. $-2 1/3$ and $-3$
- 4. $2 1/3$ and $-3$
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The cross method is a factorisation method used for quadratics.

Since the equation is in standard form $($ $a$ $x^2+$ $b$ $x+$ $c$ $=0)$ we can factorise using the cross method.

$3$ $x^2$ $+2$ $x$ $-21$ $=0$

Find two numbers that multiply to $3$ on the left side and multiply to $-21$ on the right side. Also, the coefficients of $x$ in the factors, when multiplied by the constants, should add to $2x$ .

$3$ and $1$ fit the condition for the left side and $-7$ and $3$ fit the condition for the right side

$3 xx 1$ $=$ $3$

$-7 xx 3$ $=$ $-21$

$(3x xx 3) + (x xx -7)$ $=$ $2x$

Read across to get the factors.

$(3x-7)(x+3)=0$

Solve for each value of $x$ .

$3x-7$ $=$ $0$

$3x-7$ $+7$ $=$ $0$ $+7$

$3x$ $=$ $7$

$x$ $=$ $7/3$

$x$ $=$ $2 1/3$

$x+3$ $=$ $0$

$x+3$ $-3$ $=$ $0$ $-3$

$x$ $=$ $-3$

$x=2 1/3, -3$
Question 4 of 4

4. Question
Solve for $x$ :

$-2x^2-11x-12=0$
- 1. $3/2$ and $4$
- 2. $-3/2$ and $-4$
- 3. $-3/2$ and $4$
- 4. $3/2$ and $-4$
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The cross method is a factorisation method used for quadratics.

Since the equation is in standard form $($ $a$ $x^2+$ $b$ $x+$ $c$ $=0)$ we can factorise using the cross method.

$-2$ $x^2$ $-11$ $x$ $-12$ $=0$

Find two numbers that multiply to $-2$ on the left side and multiply to $-12$ on the right side. Also, the coefficients of $x$ in the factors, when multiplied by the constants, should add to $-11x$ .

$-2$ and $1$ fit the condition for the left side and $-3$ and $4$ fit the condition for the right side

$-2 xx 1$ $=$ $-2$

$-3 xx 4$ $=$ $-12$

$(-2x xx 4) + (x xx -3)$ $=$ $-11x$

Read across to get the factors.

$(-2x-3)(x+4)=0$

Solve for each value of $x$ .

$-2x-3$ $=$ $0$

$-2x-3$ $+3$ $=$ $0$ $+3$

$-2x$ $=$ $3$

$x$ $=$ $-3/2$

$x+4$ $=$ $0$

$x+4$ $-4$ $=$ $0$ $-4$

$x$ $=$ $-4$

$x=-3/2, -4$

Solve Quadratics by Factoring

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Quiz summary

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2. Question

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3. Question

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4. Question

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Quizzes

$- 5 + (- 1)$ $-5 + (-1)$	$=$ $=$	$- 6$ $-6$
$- 5 \times - 1$ $-5 xx -1$	$=$ $=$	$5$ $5$

$x - 5$ $x-5$	$=$ $=$	$0$ $0$
$x - 5$ $x-5$ $+ 5$ $+5$	$=$ $=$	$0$ $0$ $+ 5$ $+5$
$x$ $x$	$=$ $=$	$5$ $5$

$x - 1$ $x-1$	$=$ $=$	$0$ $0$
$x - 1$ $x-1$ $+ 1$ $+1$	$=$ $=$	$0$ $0$ $+ 1$ $+1$
$x$ $x$	$=$ $=$	$1$ $1$

$2 \times 1$ $2 xx 1$	$=$ $=$	$2$ $2$
$- 1 \times - 1$ $-1 xx -1$	$=$ $=$	$1$ $1$
$(2 x \times - 1) + (x \times - 1)$ $(2x xx -1) + (x xx -1)$	$=$ $=$	$- 3 x$ $-3x$

$2 x - 1$ $2x-1$	$=$ $=$	$0$ $0$
$2 x - 1$ $2x-1$ $+ 1$ $+1$	$=$ $=$	$0$ $0$ $+ 1$ $+1$
$2 x$ $2x$	$=$ $=$	$1$ $1$
$x$ $x$	$=$ $=$	$\frac{1}{2}$ $1/2$

$3 xx 1$	$=$	$3$
$-7 xx 3$	$=$	$-21$
$(3x xx 3) + (x xx -7)$	$=$	$2x$

$3x-7$	$=$	$0$
$3x-7$ $+7$	$=$	$0$ $+7$
$3x$	$=$	$7$
$x$	$=$	$7/3$
$x$	$=$	$2 1/3$

$-2 xx 1$	$=$	$-2$
$-3 xx 4$	$=$	$-12$
$(-2x xx 4) + (x xx -3)$	$=$	$-11x$

$-2x-3$	$=$	$0$
$-2x-3$ $+3$	$=$	$0$ $+3$
$-2x$	$=$	$3$
$x$	$=$	$-3/2$