*Help With Higher Level Maths Please?*

purechancer

Hi
Ive just started into higher level maths and today we had homework which i really didnt understand and was woundering if any one can explain this to me?

Q1]
If x+p is a factor of ax(squared) +bx+3 show that p(b-ap)=3

Q2]
If x(squared)+p is a factorof ax(cubed)+x(squared)+kx+b show that k=ap and b=p

Thanks

Ditzie

you divide the factor into the equation then when you get as far as you can you let it equal to 0 as if it's a factor it will divide in equally and you will get 0 anyway.
sorry if you don't understand what I'm trying to say, it's kinda complicated to explain without being able to do it out...

so x+p divided into ax2+bx+3.... x into ax2 is ax, then multiply by x+p to give ax2+pax... i've tried to do it out here if it's easier to understand



hahaax+(b-pa)
x+p l ax2+bx+3
haha-ax2-pax
hahax(b-pa) +3
haha-x(b-pa)-p(b-pa)

because it's equal to 0 then -p(b-pa) must be equal to 3! hope that helps

purechancer

Im still not sure i get it ive done it out a couple times and i get a different answer everytime and none of them have been right :P
Thanks anyhow though

Ditzie

sorry, i've tried to do it out but it's hard to explain through a computer, I'e probably confused you more now

JamesJB

www.studentxpress.ie

If you scroll down you can go to the HL exam papers, with solutions. Many factor problems with little proof type questions, plus an explanation for each question. IIRC the two main ways to cover these problems are found here with a bit of commentary on both :cool:

purechancer

Haha i think i get where your coming from:)
Thanks a million

Ditzie

no problem, I'm repeating maths so the revision helps but if you're still confused go to www.examinations.ie, it's the official exam committees website, they have all the marking schemes

Maybe_Memories

It's easier to equate coefficients.
Take your factor, multiply it by whatever you need to make the expression. In the first one, multiply (x + p) by (ax + k), where k is a constant.

Multiply them out, equate the two expressions, equate the coefficients and solve the equations.

Ditzie

Maybe_Memories wrote: »

It's easier to equate coefficients.
Take your factor, multiply it by whatever you need to make the expression. In the first one, multiply (x + p) by (ax + k), where k is a constant.

Multiply them out, equate the two expressions, equate the coefficients and solve the equations.

funny I'd find that more confusing!

JamesJB

Maybe_Memories wrote: »

It's easier to equate coefficients.
Take your factor, multiply it by whatever you need to make the expression. In the first one, multiply (x + p) by (ax + k), where k is a constant.

Multiply them out, equate the two expressions, equate the coefficients and solve the equations.

I say learn both the algebraic division method and the identity method, unless one is really confusing. You can never have too many ways to approach a problem and sometimes which one is 'better' depends on the situation :cool:

Ditzie wrote: »

funny I'd find that more confusing!

Try to understand why forming an identity works. It's sort of like working backwards from the factors. Say you had x^2 + 4x + 4. Factorise it and you get (x+2)(x+2). Now let's just say you were told it was x^2 +4x + 4 and one of the factors was (x+2). The quadratic is just the product of the two linears (the factors), (x+2)(x+2). Even though the result is pretty obvious, let a stand for the other root. Then you get (x+2)(x+a). Equate them. x^2 + x(a+2) + 2a = x^2 + 4x + 4. Then you can see that a = 2, so the other factor is (x+2).

This will work for more complex problems, assuming I explained it correctly! It might be the easiest method when you work with cubic equations, too, since you're often given a factor and a quadratic with an unknown, along with the original equation. Since a cubic is a linear multiplied by a quadratic, you can multiply them out and form an identity just the same.

I'll again point to www.studentxpress.ie for a much better and more concrete explanation than this one, but I hope it helped nonetheless

MathsManiac

Alternatively, for question 1, since x+p is a factor, it follows that (-p) is a root.

So a(-p)^2 + b(-p) + 3 = 0.

If you rearrange that, you get the answer.

This only works when you have been given linear factors, so it won't work for question 2.

Personally, I'd do the second one by Maybe_Memories' method. The only way for a quadratic expression to be a factor of a cubic, is if the other factor is linear. The first term has to be ax, to make the cube term work out, and the other term has to be b/p, to make the constant term work out.

So now you know that it's (x^2 + p)(ax + b/p). Multiply that out and equate the coefficients, and Bob's yer uncle.