Engineering Polynomial - Help needed!

If you can, please also help me step by step doing the next question so that I begin to understand! Bear in mind I haven't done A level math

Calculate the roots of the following polynomial:

v^3 - v^2 - 30v + 72 = 0 given v = 4 is a root.

I can get to P(x) = (v-4)(v^2+av+b) = 0

Any help is appreciated !!!

If you can, please also help me step by step doing the next question so that I begin to understand! Bear in mind I haven't done A level math

Calculate the roots of the following polynomial:

v^3 - v^2 - 30v + 72 = 0 given v = 4 is a root.

For C. The left hand side is simply your original P(x). The right hand side is B, where they have grouped the terms together depending on the exponent of x.

So, ax^2 - 4x^2 becomes (a-4)x^2. Similarly the other terms.

I'm being stupid but why in the example is it + + - ?

x³ + (a-4)x² + (B-4a)x - 4b

You mean:
$P(v) = (v-4)(v^2+\alpha v+ \beta)$ not $P(x)$

Full solutions aren't really done here , but I'll see what I can do.
OK, so you know that:
$P(v) = (v-4)(v^2+\alpha v+ \beta) = v^3 - v^2 - 30v + 72$
From here, you know that the left hand side must equal the right hand side, so there must be:

Since you know that $\alpha , \beta$ are some constants, you can solve for them, and hence find the quadratic, if that makes sense?

Helping hand - only use if absolutely necessary.

I'm teaching myself A Level Maths/Engineering Mathematics for an engineering degree that I have been accepted for 13/14 (As I didn't take A level maths..).

I've got stuck at the polynomial equation and how to work out all the roots when given one (I'm using Engineering Mathematics by Croft and Davison).

I have passed the quadratics stage and can use all the methods to solve those equations, however looking at the next part of the book i.e polynomials I have gotten quite stuck! Even the example doesn't count as I don't understand what is really going on




I've labelled the parts I'm stuck at - ABC.

How do you get from equation B to C ? And how do you know which way around the +/- signs go as I can't figure it out?

If you can, please also help me step by step doing the next question so that I begin to understand! Bear in mind I haven't done A level math

Calculate the roots of the following polynomial:

v^3 - v^2 - 30v + 72 = 0 given v = 4 is a root.

I can get to P(x) = (v-4)(v^2+av+b) = 0

Any help is appreciated !!!

Putting all the signs in, we actually have in B.

+ax^2 and -4x^2, hence + (+a-4)x^2 = +(a-4)x^2

and +bx and -4ax becomes +(+b-4a)x = +(b-4a)x

IF we had:

-bx -4ax we would have had + (-b-4a)x = - (b+4a) x.

Hope that helps, as I've not addressed your question directly.

I'm not quite sure I'm just trying to follow the book method ^^ so I did this:

$v^3-v^2-30v+72=0$

Given

v=4 therefore (v-4)

$p(v)=v^3-v^2-30v+72=(v-4)(v^2+av+b)[br][br]p(v)=v^3-v^2-30v+72=v^3+av^2+bv-4v^2-4av-4b=0[br][br]p(v)=v^3-v^2-30v+72=v^3+(a-4)v^2+(b-4a)v-4b[br][br]72=-4b [br]b=-18[br][br]1=a-4[br]a=5$

Is this heading the right way?

Ahhh Indeed that does help ! Doesn't that mean that it should always be ++- ?

Looking at the last few lines in that textbook and seeing them equate two irrational numbers to two finite decimals without so much as an "$\approx$" sign gives me this horrible uneasy feeling :| *runsoff*

I've not checked, but it looks good.
Another way to check is to expand out the brackets.

I'm not sure what that is, so I will try and look it up.

From what I wrote above I get:

$(v-4)(v^2+5v-18)=0$

Which can be put into the quad formula, however when I do this I get the roots of v at 4, -2.4245 and 7.4245...

however the answers are supposed to be 4, 3 and -6..

ermmm

Spotted your mistake.
It should read
$-1=a-4$

Well now I'm really lost

Well the textbook is wrong - sickeningly so - though it doesn't really matter for Engineering purposes

God dam it!! Thank you so much for your help - In the meantime I'm gonna try out the next question just to double check I'm doing things right:

$2y^3+3y^2-11y+3=0$

Given y = 1.5 is a root.

EDIT: Just saw the 2y^3.. will try figure what the 2 will do (Think it may just effect the quad formula but not anything else ?)

Spotted your mistake.
It should read
$-1=a-4$

Okay I ended up with the wrong answers again lol. I'll write what I did:

$2y^3+3y^2-11y+3=0[br][br][br]p(y)=(y-1.5)(y^2+ay+b)[br][br][br]p(y)=y^3+ay^2+by-1.5y^2-1.5ay-1.5b[br][br][br]3=-1.5b[br][br][br]b=-2[br][br][br]3=a-1.5[br][br][br]a=4.5[br][br][br]x=(-4.5+/-sqrt((4.5)^2-4(-2)))/2$

This gives me x = 0.4076 or x = -4.9076 (should be -3.3128 or 0.3098)

I think as its $2y^3$ I may need to do something different ?