Free Printable Factorising and solving quadratics

1.

Factorise

𝑥^2+7𝑥+12

Step 1: What two numbers multiply to get 12, and add up to get 7?

a)

2

b)

3

c)

4

d)

6

2.

Factorise

Step 2: Given that the two numbers were 3 and 4, what is the correct factorisation of

𝑥^2+7𝑥+12

a)

$3\left(x+4\right)$

b)

$4\left(x+3\right)$

c)

$x\left(3+4\right)$

d)

$\left(x+3\right)\left(x+4\right)$

3.

Solve

𝑥^2+7𝑥+12=0

Step 3: Given that the factorisation of the above is

\left(x+3\right)\left(x+4\right)=0

What are the two possible possible solutions for x?

a)

$x=3$

b)

$x=-3$

c)

$x=4$

d)

$x=-4$

4.

Factorise

𝑥^2+10𝑥+25

Step 1: What two numbers multiply to get 10, and add up to get 25?

a)

3

b)

5

c)

5

d)

7

5.

Factorise

Step 2: Given that the two numbers were 5 and 5, what is the correct factorisation of

𝑥^2+10𝑥+25

a)

$5\left(x+5\right)$

b)

$x\left(5+5\right)$

c)

$\left(x+5\right)\left(x+5\right)$

6.

Solve

𝑥^2+10𝑥+25=0

Step 3: Given that the factorisation of the above is

\left(x+5\right)\left(x+5\right)=0

What is the only possible possible solution for x?

a)

$x=5$

b)

$x=-5$

7.

Factorise

𝑥^2+8𝑥+12

Step 1: What two numbers multiply to get 12, and add up to get 8?

a)

2

b)

3

c)

4

d)

6

8.

Factorise

Step 2: Given that the two numbers were 2 and 6, what is the correct factorisation of

𝑥^2+8𝑥+12

a)

$2\left(x+6\right)$

b)

$6\left(x+2\right)$

c)

$x\left(2+6\right)$

d)

$\left(x+2\right)\left(x+6\right)$

9.

Solve

𝑥^2+8𝑥+12=0

Step 3: Given that the factorisation of the above is

\left(x+2\right)\left(x+6\right)=0

What are the two possible possible solutions for x?

a)

$x=2$

b)

$x=-2$

c)

$x=6$

d)

$x=-6$

10.

Solve and Factorise

𝑥^2+8𝑥+15=0

Step1: Factorise

𝑥^2+8𝑥+15

a)

$\left(x+2\right)\left(x+6\right)$

b)

$\left(x+3\right)\left(x+5\right)$

c)

$\left(x+1\right)\left(x+8\right)$

d)

$\left(x+4\right)\left(x+4\right)$

11.

Solve and factorise

𝑥^2+8𝑥+15=0

Step 2: given that the factorisation of the above is

\left(x+3\right)\left(x+5\right)=0

What are the two possible solutions for x?

a)

$x=3$

b)

$x=-3$

c)

$x=5$

d)

$x=-5$

12.

Solve and Factorise

𝑥^2+11𝑥+28=0

Step1: Factorise

𝑥^2+11𝑥+28

a)

$\left(x+3\right)\left(x+7\right)$

b)

$\left(x+4\right)\left(x+7\right)$

c)

$\left(x+5\right)\left(x+6\right)$

d)

$\left(x+2\right)\left(x+9\right)$

13.

Solve and factorise

𝑥^2+11𝑥+28=0

Step 2: given that the factorisation of the above is

\left(x+4\right)\left(x+7\right)=0

What are the two possible solutions for x?

a)

$x=4$

b)

$x=-4$

c)

$x=7$

d)

$x=-7$

14.

Solve and Factorise

𝑥^2+3𝑥−18=0

Step1: Factorise

𝑥^2+3𝑥−18

a)

$\left(x+3\right)\left(x-6\right)$

b)

$\left(x-3\right)\left(x+6\right)$

c)

$\left(x+2\right)\left(x-9\right)$

d)

$\left(x-2\right)\left(x+9\right)$

15.

Solve and factorise

𝑥^2+3𝑥−18=0

Step 2: given that the factorisation of the above is

\left(x-3\right)\left(x+6\right)=0

What are the two possible solutions for x?

a)

$x=3$

b)

$x=-3$

c)

$x=6$

d)

$x=-6$

Factorising and solving quadratics

Total questions: 15

Worksheet time: 30mins

Similar Resources on Wayground