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Solving Equations by Factoring

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Solving Equations by Factoring 1

Solving Equations by Factoring 1

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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 $- 3$ $-3$
- 3. $- 2 \frac{1}{3}$ $-2 1/3$ and $- 3$ $-3$
- 4. $- 2 \frac{1}{3}$ $-2 1/3$ and $3$ $3$
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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.

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

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

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

$3 \times 1$ $3 xx 1$ $=$ $=$ $3$ $3$

$- 7 \times 3$ $-7 xx 3$ $=$ $=$ $- 21$ $-21$

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

Read across to get the factors.

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

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

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

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

$3 x$ $3x$ $=$ $=$ $7$ $7$

$x$ $x$ $=$ $=$ $\frac{7}{3}$ $7/3$

$x$ $x$ $=$ $=$ $2 \frac{1}{3}$ $2 1/3$

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

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

$x$ $x$ $=$ $=$ $- 3$ $-3$

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

4. Question
Solve for $x$ $x$ :

$- 2 x^{2} - 11 x - 12 = 0$ $-2x^2-11x-12=0$
- 1. $\frac{3}{2}$ $3/2$ and $- 4$ $-4$
- 2. $- \frac{3}{2}$ $-3/2$ and $4$ $4$
- 3. $\frac{3}{2}$ $3/2$ and $4$ $4$
- 4. $- \frac{3}{2}$ $-3/2$ and $- 4$ $-4$
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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.

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

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

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

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

$- 3 \times 4$ $-3 xx 4$ $=$ $=$ $- 12$ $-12$

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

Read across to get the factors.

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

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

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

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

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

$x$ $x$ $=$ $=$ $- \frac{3}{2}$ $-3/2$

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

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

$x$ $x$ $=$ $=$ $- 4$ $-4$

$x = - \frac{3}{2}, - 4$ $x=-3/2, -4$

Quizzes

Sum & Product of Roots 1
Sum & Product of Roots 2
Sum & Product of Roots 3
Sum & Product of Roots 4
Solving Equations by Factoring 1
Solving Equations Using the Quadratic Formula
Completing the Square 1
Completing the Square 2
Intro to Quadratic Functions (Parabolas) 1
Intro to Quadratic Functions (Parabolas) 2
Intro to Quadratic Functions (Parabolas) 3
Graph Quadratic Functions in Standard Form 1
Graph Quadratic Functions in Standard Form 2
Graph Quadratic Functions by Completing the Square
Graph Quadratic Functions in Vertex Form
Write a Quadratic Equation from the Graph
Write a Quadratic Equation Given the Vertex and Another Point
Quadratic Inequalities 1
Quadratic Inequalities 2
Quadratics Word Problems 1
Quadratics Word Problems 2
Quadratic Identities
Graphing Quadratics Using the Discriminant
Positive and Negative Definite
Applications of the Discriminant 1
Applications of the Discriminant 2
Solving Reducible Equations

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