Engineering Polynomial - Help needed! Watch

clloyd12
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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 !!!
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joostan
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(Original post by clloyd12)

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 !!!
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:
  • 1v^3 on both sides.
  • -1v^2 on both sides.
  • -30v on both sides.
  • and 72 on both sides.

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.
Spoiler:
Show
How to solve for \beta I'll let you figure out how to solve for \alpha
Spoiler:
Show
(v-4)(v^2+\alpha v+ \beta) = v^3 - v^2 - 30v + 72

\Rightarrow (-4)(\beta) = 72

\Rightarrow \beta = -18
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ghostwalker
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(Original post by clloyd12)
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?

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.

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.
This is an identical method to the first question. So, make sure you understand that example fully first, then look at the second question.
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clloyd12
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(Original post by ghostwalker)
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
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ghostwalker
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(Original post by clloyd12)
I'm being stupid but why in the example is it + + - ?

x3 + (a-4)x2 + (B-4a)x - 4b
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.
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clloyd12
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(Original post by joostan)
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:
  • 1v^3 on both sides.
  • -1v^2 on both sides.
  • -30v on both sides.
  • and 72 on both sides.

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.
Spoiler:
Show
How to solve for \beta I'll let you figure out how to solve for \alpha
Spoiler:
Show
(v-4)(v^2+\alpha v+ \beta) = v^3 - v^2 - 30v + 72

\Rightarrow (-4)(\beta) = 72

\Rightarrow \beta = -18
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)



p(v)=v^3-v^2-30v+72=v^3+av^2+bv-4v^2-4av-4b=0



p(v)=v^3-v^2-30v+72=v^3+(a-4)v^2+(b-4a)v-4b



72=-4b 

b=-18



1=a-4

a=5

Is this heading the right way?
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theterminator
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(Original post by clloyd12)
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 !!!
I don't know if you are but just in case you are missing the obvious:

Any polynomial can be written as the expansion of brackets. e.g. 2nd degree polynomial:
 Ax^2 + Bx + C = (ax+b)(cx+d)

\newline

expanding:

\newline

= acx^2 + x(ad + bc) + bd

\newline

compare coefficients:

\newline A = ac, B = ad+bc, C = bd

\newline

If given A,B,C, you can find the corresponding terms a,b,c,d to get the bracket form. 

\newline

This can be done with 3rd, 4th, etc... degree polynomials also
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clloyd12
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(Original post by ghostwalker)
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.
Ahhh Indeed that does help ! Doesn't that mean that it should always be ++- ?
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joostan
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(Original post by clloyd12)
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)



p(v)=v^3-v^2-30v+72=v^3+av^2+bv-4v^2-4av-4b=0



p(v)=v^3-v^2-30v+72=v^3+(a-4)v^2+(b-4a)v-4b



72=-4b 

b=-18



1=a-4

a=5

Is this heading the right way?
I've not checked, but it looks good.
Another way to check is to expand out the brackets.
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ghostwalker
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(Original post by clloyd12)
Ahhh Indeed that does help ! Doesn't that mean that it should always be ++- ?
If your breakdown is in this form "(v-4)(v^2+av+b)", yes. But it could be, in another example, "(x+7)(x^2+ax+b)" in which case they'll all be positive, or something else in a different variety of question.

Don't be concerned with the + + - form, but rather understand how they have arisen here. Otherwise you'll end up with lots of little rules depending on different circumstances, whereas you're just doing basic algebraic manipulation.
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Jkn
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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*
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clloyd12
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(Original post by Jkn)
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*
Well now I'm really lost
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clloyd12
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(Original post by joostan)
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
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joostan
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(Original post by clloyd12)
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
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clloyd12
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(Original post by joostan)
Spotted your mistake.
It should read
-1=a-4
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 ?)
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Jkn
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(Original post by clloyd12)
Well now I'm really lost
Well the textbook is wrong - sickeningly so - though it doesn't really matter for Engineering purposes
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clloyd12
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(Original post by Jkn)
Well the textbook is wrong - sickeningly so - though it doesn't really matter for Engineering purposes
For someone that didn't do a level maths and is trying to self teach engineering maths I have no idea
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joostan
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(Original post by clloyd12)
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 ?)
No problem
Hint:
Spoiler:
Show
When you see a non integer root, try and make it an integer.
So if (y-1.5) is a root of f(y) then so is (2y-3)
Then follow the same process.
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clloyd12
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(Original post by joostan)
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





p(y)=(y-1.5)(y^2+ay+b)





p(y)=y^3+ay^2+by-1.5y^2-1.5ay-1.5b





3=-1.5b





b=-2





3=a-1.5





a=4.5





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 ?

EDIT: Just saw your post above about (2y - 3) Will try figure it out now ^^
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joostan
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(Original post by clloyd12)
Okay I ended up with the wrong answers again lol. I'll write what I did:

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





p(y)=(y-1.5)(y^2+ay+b)





p(y)=y^3+ay^2+by-1.5y^2-1.5ay-1.5b





3=-1.5b





b=-2





3=a-1.5





a=4.5





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 ?
It's 2y^3
As such, when you equate the coefficients of y^3 there needs to be 2 on each side.
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