Year 12s: what will be the first way you research your future options? >>

confusing question

WWEKANE

how can you work out the remaining part without expanding?

Reply 1

username3314392

Could you type the question?

Reply 2

WWEKANE

a square hole of side length a+2 is cut from a larger square of side length 2a+5. without expanding ANY brackets write the remaining part of the the larger square as a pair of factors

IMG_20181103_132155-compressed.jpg.jpeg111.3KB

Reply 3

WWEKANE

the answer on the mark scheme is (a+3)(3a+7)

Reply 4

3pointonefour

The larger square has area (2a+5)^2
Smaller square has area (a+2)^2

How do you simplify (2a+5)^2 - (a+2)^2?

HINT:

Spoiler

Reply 5

student322

Original post by WWEKANE

a square hole of side length a+2 is cut from a larger square of side length 2a+5. without expanding ANY brackets write the remaining part of the the larger square as a pair of factors

is this from gcse higher?? what exam board??

Reply 6

WWEKANE

no it is from my as level book chapter 2

Reply 7

hiyatt

(Original post by WWEKANE) IMG_20181103_132155-compressed.jpg.jpeg

how can you work out the remaining part without expanding?
use the difference of two squares.
That says that for a and b any real number we have

a^2-b^2=(a+b)(a-b)

Reply 8

3pointonefour

Original post by hiyatt

(Original post by WWEKANE) IMG_20181103_132155-compressed.jpg.jpeg

how can you work out the remaining part without expanding?
use the difference of two squares.
That says that for a and b any real number we have

a^2-b^2=(a+b)(a-b)

a and b don't necessarily have to be real numbers

Reply 9

hiyatt

Original post by 3pointonefour

a and b don't necessarily have to be real numbers

True, but this question only involves real numbers

Reply 10

WWEKANE

I'm not sure if I've done this right but I expanded each term by the power then factorised

(2a+5)(2a-5) and (a+2)(a-2) what do I do from here

Reply 11

hiyatt

Original post by WWEKANE

I'm not sure if I've done this right but I expanded each term by the power then factorised

(2a+5)(2a-5) and (a+2)(a-2) what do I do from here

This is not really necassary. So,

x^2 - y^2 = (x+y)(x-y)

is what is known as an identity. No matter what real numbers you substitute for x and y this will always hold.

if you look at your expression:

(2a+5)^2 - (a+2)^2

what you notice is that you have (if you look bac at the identity) x replaced with 2a+5 and y replaced with a+2 on the left hand side of that identity. Whatever a you substitute into your expression, you get the difference of two real numbers so what this tells you is that you can substitute (2a+5) for x on the rhs and (a+2) for y on the right hand side knowing that whatever real number you subsitute for a it will be the same as your original expression