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Improving algebra skills.

Does anyone know where I can start?

It seems as though I've coasted through GCSEs and AS and reached A2 maths with poor algebra skills. Just due to neglect or indifference in teaching, not due to lack of ability. (I suppose I'm fairly good with maths outside of algebra, but my skills for this particular topic are indeed quite sub-par.)

I appreciate that in order to reach my goals of an A, and eventually 90%+, that I'd need to improve my algebra skills a fair bit.

I'll quite happily do so, that's not a problem.

So with this thread, I begin.

Suppose I have a question such as differentiate: y=(2x-1)³(x+1)³

Fairly basic right?

u=(2x-1)³, du/dx = 6(2x-1)²
v=(x+1)³, dv/dx = 3(x+1)²
dy/dx = (x+1)³*6(2x-1)²+ (2x-1)³*3(x+1)²

Here's the first hiccup in my thought process.

It just looks like a mess to work with?
What do I do now?
What factor do I take out?
Is there anyway of quick-fire determining the factor to take?
A thought checklist one can keep to find so?

Ultimately, my goal is to factorise the expression found above.

Here's where I need help:

I need to fix my thought process, every step preceding such has been an almost systematic, intuitive thought pattern.

1) Sort into u and x.
2)Differentiate above respectively.
3)Plug into the product rule.
4).....?

Can anyone share their thoughts step by step going through this problem?

Especially when factorising, as that's a major pitfall for me.

Thanks in advance!

Reply 1

BabyMaths

Original post by MindArchitect

dy/dx = (x+1)³*6(2x-1)²+ (2x-1)³*3(x+1)²

Especially when factorising, as that's a major pitfall for me.

Thanks in advance!

As a first step in factorising any expression take out common factors.

Here your common factors are there to see. You have a number, a power of x+1 and a power of 2x-1.

?(x+1)^?(2x-1)^?(.......)

Reply 2

MindArchitect

Original post by BabyMaths

As a first step in factorising any expression take out common factors.

Here your common factors are there to see. You have a number, a power of x+1 and a power of 2x-1.

?(x+1)^?(2x-1)^?(.......)

I did indeed find them as such, and did the following step:

3(x+1)²(2x-1)²[2(x+1)+2x-1]

3(X+1)²(2x-1)²[4x+1]

See, I know this is right but don't know why, mainly because I don't see what multiplies together to give 6(2x-1)² from the original unfactorised expression.

Does the 3 multiply both Quadratics outside []? Well of-course it does, so the 3 must multiply by a 2 to give the 6.

If so, what gives 6(2x-1)²when expanding?

Care to to explain?

Only thing I can think of at present is that my working is wrong.

(edited 9 years ago)

Reply 3

TenOfThem

Original post by MindArchitect

dy/dx = (x+1)³*6(2x-1)²+ (2x-1)³*3(x+1)²

Original post by MindArchitect

See, I know this is right but don't know why, mainly because I don't see what multiplies together to give 6(2x-1)² from the original unfactorised expression.

(x+1)^3\times 6(2x-1)^2 + (2x-1)^3\times 3(x+1)^2 = 6(x+1)^3(2x-1)^2 + 3(x+1)^2(2x-1)^3

Because you can multiply in any order 3x4 = 4x3 or 2x3x5 = 2x5x3 = 3x2x5 etc

So now you have

2*3(x+1)(x+1)(x+1)(2x-1)(2x-1) + 3(x+1)(x+1)(2x-1)(2x-1)(2x-1)

=

3(x+1)^2(2x-1)^2[2(x+1) + (2x-1)]

(edited 9 years ago)

Reply 4

davros

Original post by arkanm

Learn vedic maths, it's brilliant. For example, to solve a quadratic, say

ax^2+bx+c

, you differentiate it and set it equal to

\pm

the square root of the disciminant like this:

2ax+b=\sqrt{b^2-4ac}

.

This makes it much easier to remember how to solve a quadratic. Of course, rearranging it gives the quadratic formula.

Not sure if you're bored, or what, but please stop trolling other people's thread's with irrelevant suggestions. The OP is stuck on factorisation of complex algebraic expressions, not solving quadratics!