The Student Room Group

AS maths factorization help

Never thought I'd be struggling with quadratics, but I've never seen a question like this. Anyone have a link to a video explaining this method?

(question posted below).
Original post by quickquestion805
Never thought I'd be struggling with quadratics, but I've never seen a question like this. Anyone have a link to a video explaining this method?

(question posted below).

Question:
fact1.png
Answer:

fact2.png
Reply 2
Original post by quickquestion805
Question:
fact1.png
Answer:

fact2.png

what's the problem? They've pretty much done it by inspection - spotting that there's a 4x-y at the end and also that the 1st 2 terms are just y(4x-y), hence 4x-y can be taken out as a factor of the entire expression.
Reply 3
Original post by quickquestion805
Question:
fact1.png
Answer:

fact2.png

Apart from spotting the partial factorisation in the first step (not hard), you could note that as its a quadratic the answer is going to be of the form
(x+y+1)(x+y+1)
Note Ive not put the coeffs in.
* There is no constant in the original expression, but there are linear terms. So one of the +1s is missing so
(x+y+1)(x+y)
* There is no x^2 in the original expression so one of the x's is missing. Obv it cant be the second one as there are some x's in the original expression so
(y+1)(x+y)
* Then just introduce the coeffs and match.

More "forwards", there is a y^2 so there must be a y in each factor
(y + ...)(y + ...)
Then there is an xy so
(y + x)(y + ...)
Then the last terms are x and y so that gives a constant in the second factor
(y + x)(y + 1)
Then again, introduce coeffs and ...

Another attack is to spot that y=-1 makes the original expression zero so (y+1) is a factor so
(y+1)(...)
and the other factor drops out. The answer has done something similar except they note that its zero when y=4x so the factor is (4x-y)...
(edited 11 months ago)
Original post by davros
what's the problem? They've pretty much done it by inspection - spotting that there's a 4x-y at the end and also that the 1st 2 terms are just y(4x-y), hence 4x-y can be taken out as a factor of the entire expression.

That makes sense. Interesting, I've never come across this type of question before. Guess I've gotta hit the books
Reply 5
This is a case of "trying until it works", to be quite honest.
There really is no "question type" when it comes to factorization, IMO. A lot of it is from (i) a spark of inspiration; and (ii) just doing a decent amount of exercises to get used to nice things in maths.
Reply 6
Original post by quickquestion805
That makes sense. Interesting, I've never come across this type of question before. Guess I've gotta hit the books


One "standard" but laborious way if inspiration fails is to write it as a quadratic in y so
-y^2 + (4x-1)y + 4x
Treating it as an equation = 0 (as youre looking for factors/roots) then
y = -(4x-1)/-2 +/-sqrt((4x-1)^2 + 16x)/-2
y = (4x-1)/2 +/-sqrt((4x+1)^2 )/2
y = 4x,-1
so factors (y+1)(y-4x). Then just multiply by -1 to match signs.

However, questions like this are usually constructed to make the factors/roots (values of x and/or y which make the expression equal to zero) "easy" to spot.
(edited 11 months ago)
Original post by mqb2766
One "standard" but laborious way if inspiration fails is to write it as a quadratic in y so
-y^2 + (4x-1)y + 4x
Treating it as an equation = 0 (as youre looking for factors/roots) then
y = -(4x-1)/-2 +/-sqrt((4x-1)^2 + 16x)/-2
y = (4x-1)/2 +/-sqrt((4x+1)^2 )/2
y = 4x,-1
so factors (y+1)(y-4x). Then just multiply by -1 to match signs.

However, questions like this are usually constructed to make the factors/roots (values of x and/or y which make the expression equal to zero) "easy" to spot.


Yeah that is quite laborious. I'm sure I could've figured out the mark scheme method, but I was more surprised to see such an unfamiliar factorization question in a textbook that rarely varies from the same old stuff. Thanks for the help

Quick Reply

Latest