Question 1

LEVEL 4

Factorise the following into a single bracket:

10x+20

Select the correct answer from the list below:

A: 10(x+2)

B: 10(x+20)

C: 5(2x+4)

D: 4(2x+5)

 

Correct answer: A: 10(x+2)

Worked solution:

To factorise you need to take the highest factor outside of the bracket,

leaving the other factors inside. The easiest way to do this is see what is

the largest number that goes into 10 and 20, which is 10.

10 goes outside of the bracket leaving x+2 inside.

You can always check to see if you have factorised correctly, by multiplying

out and seeing if you get back to the original expression.

Question 2

LEVEL 4

Factorise the following into a single bracket:

12x+10

Select the correct answer from the list below:

A: 4(3x+2.5)

B: 2(6x-5)

C: 2(6x+5)

D:2(5x+6)

 

Correct answer: C: 2(6x+5)

Worked solution:

To factorise you need to take the highest factor outside of the bracket,

leaving the other factors inside. The easiest way to do this is see what is

the largest number that goes into 12 and 10, this is 2.

2 goes outside of the bracket leaving 6x+5 inside.

You can always check to see if you have factorised correctly, by multiplying

out and seeing if you get back to the original expression.

Question 3

LEVEL 4

Factorise the following into a single bracket:

3x^2 + 12x

Select the correct answer from the list below:

A: 3x^2(x+4)

B: 3x(x+12)

C:  x(3x+12)

D: 3x(x+4)

 

Correct answer: D: 3x(x+4)

Worked solution:

To factorise you need to take the highest factor outside of the bracket,

leaving the other factors inside. The easiest way to do this is see what is

the largest number and letter that goes into

3x^2 and 12x

Therefore 3x goes outside of the bracket leaving (x+4) inside.

Question 4

LEVEL 4

Factorise the following into a single bracket:

64x^2 + 28x

Select the correct answer from the list below:

A: x(64x+28)

B: 8x(8x+4)

C: 8x(8x+3)

D: 4x(16x+7)

 

Correct answer: D: 4x(16x+7)

Worked solution:

To factorise you need to take the highest factor outside of the bracket,

leaving the other factors inside. The easiest way to do this is see what is

the largest number and letter that goes into

64x^2 and 28x

Therefore 4x goes outside of the bracket leaving (16x+7) inside.

Question 5

LEVEL 4

Factorise the following into a single bracket:

15ab + 25abc.

Select the correct answer from the list below:

A: 5(3ab+5abc)

B: 5ab(3+5c)

C: -10abc

D: -10c

 

CORRECT ANSWER: B 5ab(3+5c)

WORKED SOLUTION:

We start by finding the highest factor of the numbers (10 and 25)

which is 5, so we can take that out as a factor.

15ab+25abc =5(3ab+5abc)

And then we look at the terms inside the bracket

and see if there are any common variables (letters).

Here they both have an a of the same power, so we can pull that out.

15ab-25abc =5a(3b+5bc)

And doing this step again we can see that both have

a b of the same power, so we can pull that out as well.

15ab-25abc =5ab(3+5c)

And we can’t do anymore, so this is factorised fully.

Question 6

LEVEL 4

Factorise fully 16x^2y^2+4xy.

Select the correct answer from the list below:

A: 20x^3y^3

B: 4xy(xy+1)

C: 4(4xy+1)

D: 4xy(4xy+1)

 

CORRECT ANSWER: D 4xy(4xy+1)

WORKED SOLUTION:

We start by finding the highest factor of the numbers (16 and 4) which is 4, so we can take that out as a factor.

16x^2y^2+4xy=4(4x^2y^2+xy)

And then we look at the terms inside the bracket and see if there are any common variables (letters). Here they both have an x. Because one is an x^2 and the other is x=x^1, we have to take out x=x^1 because it is smallest.

16x^2y^2+4xy=4x(4xy^2+y)

And doing this step again we can see that they both have a y. Because one is a y^2 and the other is y=y^1, we have to take out y=y^1 because it is smallest.

16x^2y^2+4xy=4xy(4xy+1)

And we can’t do anymore, so this is factorised fully.

Question 7

LEVEL 4

Factorise fully 2abc^2-2ab^2+4ab .

Select the correct answer from the list below:

A: 2ab(c^2-b+2)

B: 4c(ac-b+1)

C: 4ac^2

D: 2abc^2(b+2)

 

CORRECT ANSWER: A 2ab(c^2-b+2)

WORKED SOLUTION:

We start by finding the highest factor of the numbers (2, 2 and 4) which is 2, so we can take that out as a factor.

2abc^2-2ab^2+4ab =2(abc^2-ab^2+2ab)  

And then we look at the terms inside the bracket and see if there are any common variables (letters). Here they all have an a of the same power, so we can pull that out.

2abc^2-2ab^2+4ab =2a(bc^2-b^2+2b)  

And doing this step again we can see that they all have a b. Because one is a b^2 and the others are b=b^1, we have to take out b=b^1 because it is smallest.

2abc^2-2ab^2+4ab =2ab(c^2-b+2)

And we can’t do anymore, as only once term has a c so this is factorised fully.

QUESTION 8

LEVEL 4

Factorise fully 5xy^3-x^2y^2+4xy^3 .

Select the correct answer from the list below:

A: 5(5y^2-x^2+4y)

B: 5y-x+4y

C: xy^2(5y-x+4y)

D: 8y^4

 

CORRECT ANSWER: C xy^2(5y-x+4y)

WORKED SOLUTION:

We start by finding the highest factor of the numbers (5, 1 and 4) which is 1, so we can’t take any numbers out.

And then we look at the terms to see if there are any common variables (letters). We can see that they all have an x. Because one is an x^2 and the others are x=x^1, we have to take out x=x^1 because it is smallest.

5xy^3-x^2y^2+4xy^3 =x(5y^3-xy^2+4y^3)

Repeating this step, we can see that they all have a y. Because two are y^3 and the other is y^2, we have to take out y^2 because it is smallest.

5xy^3-x^2y^2+4xy^3 =xy^2(5y-x+4y)

And we can’t do anymore, so this is factorised fully.