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Maths C2 Factor Theorem homework help

hello, any help with questions 10 and 11 would be greatly appreciated. I have already worked out that a=-7, but I'm unsure as to how to show the equation has only one real root, and, frankly, I am completely lost on how to start question 11. Thank you in advance for any help

Reply 1

Zacken

Original post by Imo_rai

Factorise the quadratic into a linear factor and a quadratic factor, i.e: of the form (x-3)(ax^2 + bx + c) = 0, you know one solution is given by x=3 and the other two by ax^2 + bx + c = 0. So you want the latter term to have 0 solutions, what does the discriminant have to be for that to happen? Compute the discriminant and show that it is indeed that.

Reply 2

Zacken

Original post by Imo_rai

Q11: Look for factors, try plugging in x=-1, 1, -2, 2, -3, 3 - etc... look for factors, factorise the equation as much as you can.

Also compute

\frac{dy}{dx}

, which will get you a cubic, set this equal to zero - solve by factorising, try values, factorise, etc...

Then find the coordinates of the stationary points and plot them on a graph making use of the fact that you've factorised y so you know where the roots are going to be, you know the stationary points as well, that should be all.

Reply 3

XxKingSniprxX

Original post by Imo_rai

For Question 10, (x-3) = 0 in other words x = 3
Set the equation to zero, 4x^3 + ax^2 -13x + 6 = 0.
Sub x = 3 into the equation above and move everything to the other side
until you have a x^2 = constant. Divide the constant by x^2 where x = 3 and you should find the value of a.

1 real root means b^2-4ac = 0.

Q11) At a stationery point, dy/dx = 0 so differentiate the equation y = 2x^4-7x^2-6x and make it equal to zero. Rearrange dy/dx=0 to find the value of x sub that value of x back into the original equation y = 2x^4-7x^2-6x to find the coordinates (x,y). To find the nature of the stationery point, d^2y/dx^2 (double differentiate) the original equation y = 2x^4-7x^2-6x and sub the value of x in.

If d^2y/dx^2 gives a number thats negative its a max, if it gives a positive number its a minimum. If its zero then resort to first principal differentiating where you change the value of x for which dy/dx=0.

Reply 4

XxKingSniprxX

Original post by Zacken

Beat me to it. Y u no use latex? :rofl:

Is your IAL results day tomorrow? I saw you did a ton of exams in Jan = lots of results to collect tomorrow! :gasp:

Are all your STEP papers in summer I.e June?

I wish you all the best and hope you clutch up the grades. :smile:

Reply 5

Imo_rai

Original post by Zacken

Q11: Look for factors, try plugging in x=-1, 1, -2, 2, -3, 3 - etc... look for factors, factorise the equation as much as you can.

Also compute

\frac{dy}{dx}

thank you, this has really helped

Reply 6

Zacken

Original post by XxKingSniprxX

Beat me to it. Y u no use latex? :rofl:

In bed + 1:30 a.m = LaTeX some other time pls. :rofl:

Your post was more helpful though! :smile:

Is your IAL results day tomorrow? I saw you did a ton of exams in Jan = lots of results to collect tomorrow! :gasp:

It is, I'm not sure I'll be getting my results tomorrow itself because my centre is a bit useless.

Are all your STEP papers in summer I.e June?

Indeed.

I wish you all the best and hope you clutch up the grades. :smile:

Thank you very much! :smile:

Reply 7

Zacken

Original post by Imo_rai

thank you, this has really helped

Quick trick that your teacher won't know! :wink:

If you have a polynomial

x^{\text{some power}} + \cdots + d

, then if your polynomial has integer roots, you need only try values of x that divide d.

i.e: If I have

x^3 + x + 20

, the only values of x that I need try are

x = \pm 1, \pm 2, \pm 4, \pm 5, \cdots

I don't need to bother myself with

x = \pm 3

since that doesn't divide 20.

Reply 8

XxKingSniprxX

Original post by Zacken

In bed + 1:30 a.m = LaTeX some other time pls. :rofl:

Your post was more helpful though! :smile:

It is, I'm not sure I'll be getting my results tomorrow itself because my centre is a bit useless.

Indeed.

Thank you very much! :smile:

Ah, its only 9:32pm here (uk) and your +4 hours ahead of me. :redface:

Try to get some early rest if you can big day tomorrow.

My centre told me they are going to post my results home for August results day as I'm a private candidate but I told them directly I'm going to turn up and collect it as I've got adjustment etc to call up if things go good + saves a lot of stress. Is your centre doing the same or do they post it online on a website for you to see?