How did they rewrite this polynomial in this way?

[MiddleEast]

Homework Statement

Not homework

Relevant Equations

Non.

How can you rewrite polynomial in terms of (x-a) instead of x?

One thing came to mind is rewrite each x as x-a+2 (So it is x-2+2 in our example) but this will take long time and a lot of algebra steps, how did they do it very fast in the attached picture? thanks

 

[anuttarasammyak]

We can finish it as
$$f(x)=(x-2)(2x^2+x-15)=(x-2)(2x-5)(x+3)$$
Once we percieve that x=2 is a solution, we can devide the given formula by x-2.

 

[MiddleEast]

Thanks for reply. I know that, basically divide f by (x-2). But am interested in the way they did it algebraically.
Cause they gave two solutions for that problem in the book, first solution is the standard one which is dividing f by (x-2) which is a factor of f. The second solution they did as per attached image and then factor (x-2) out of three terms.

 

[FactChecker]

Homework Statement: Not homework
Relevant Equations: Non.

How can you rewrite polynomial in terms of (x-a) instead of x?

One thing came to mind is rewrite each x as x-a+2 (So it is x-2+2 in our example) but this will take long time and a lot of algebra steps, how did they do it very fast in the attached picture? thanks

Why do you think that they did it fast? Expanding x to (x-2)+2 is the the standard way to do it, as far as I know.
They probably just don't show all the (fairly routine) steps.

 

[MiddleEast]

Thanks for reply, actually I found the way.
Basically start with A(x-2) + B (x-2) + C (x-2) where A,B & C could be variables/constants.
That A must be 2x^2 to get "2x^3" in original function, that A when multiplied by -2 will give -4x^4, now you have to choose B as x so that -4x^4 & B*x will give us -3x^2 in original function. And you just continue that way. Constant term will be adjusted automatically once we get the process in right way as x-2 is factor of f.
All of this can be done in one line as they did.

 

[FactChecker]

Thanks for reply, actually I found the way.
Basically start with A(x-2) + B (x-2) + C (x-2) where A,B & C could be variables/constants.
That A must be 2x^2 to get "2x^3" in original function,

That is a logical way to proceed. But "A must be" is too strong, since A, B, and C are indistinguishable at this point. I get your idea though. Maybe you could say that the first term will include the highest power of x.

EDIT: Even then, you would have terms with a mixture of $x^n (x-2)$ factors. Are you trying to factor out (x-2)? That is not leading directly to the goal of a polynomial of $x-2$. I think it might be better to start with $A(x-2)^2+B(x-2)+C$ and see where that takes you.
But I still think that the most direct approach is to just replace all the $x$s with $(x-2)+2$ in the original polynomial and multiply it out.

 

[DaveE]

There's also the old school "long division" method I learned in junior high school:

These methods all seem essentially the same to me; each with a different look. It's just division, as you said.

 

[FactChecker]

There's also the old school "long division" method I learned in junior high school:

View attachment 347021

These methods all seem essentially the same to me; each with a different look. It's just division, as you said.

I don't see how this gives an expansion in (x-2) without several additional steps. I think there are more direct ways.

 

[DaveE]

I don't see how this gives an expansion in (x-2) without several additional steps. I think there are more direct ways.

But this is the same as the "A, B, C" method in post #5. I don't claim it's the most efficient, IDK.
What are your "more direct ways"? Surely it's not simplifying $2[(x-2)+2]^3+[(x-2)+2]-15$, is it?

 

[MatinSAR]

@MiddleEast Are you trying to derive the second polynomial equation from the first one?
$2x^3-3x^2-17x+30=2x^3-4x^2+x^2-2x-15x+30=2x^2(x-2)+x(x-2)-15(x-2)$

 

[MiddleEast]

That is a logical way to proceed. But "A must be" is too strong, since A, B, and C are indistinguishable at this point. I get your idea though. Maybe you could say that the first term will include the highest power of x.

EDIT: Even then, you would have terms with a mixture of $x^n (x-2)$ factors. Are you trying to factor out (x-2)? That is not leading directly to the goal of a polynomial of $x-2$. I think it might be better to start with $A(x-2)^2+B(x-2)+C$ and see where that takes you.
But I still think that the most direct approach is to just replace all the $x$s with $(x-2)+2$ in the original polynomial and multiply it out.

You wont start by writing A(x-2)+B(x-2)+C(x-2), you start by writing (x-2), divide first term of f by x (first term in x-2) and you would get "2x^2(x-2)", then ON SAME LINE continue writing 2x^2(x-2)+ [space] (x-2) .. now try to figure out what to put in the space by considering x^2 term coming from "2x^2(x-2)" and try to match it with coe of x^2 in original function.. continue doing this till you finish. It is basically about 30 seconds to rewrite function in terms of (x-2) so that (x-2) can be factored out. Fastest approach, knowing that x=2 is factor, is carrying out synthetic division for sure. But this way is clever and worth knowing.

 

[MiddleEast]

@MiddleEast Are you trying to derive the second polynomial equation from the first one?
$2x^3-3x^2-17x+30=2x^3-4x^2+x^2-2x-15x+30=2x^2(x-2)+x(x-2)-15(x-2)$

Not deriving, main goal is rewriting polynomial in terms of (x-a) in each term in one single step so that (x-a) will be factored out.

 

[WWGD]

Not sure if this is what you're looking for, but this is a power series about $x=2$.

 

[Frabjous]

They are also skipping the most important step of determining that 2 is a rational root.

 

[MiddleEast]

They are also skipping the most important step of determining that 2 is a rational root.

It is given in the question that x=2 is a root, and solution 1 was straightforward by dividing f by (x-2) and factor resulting quadratic equation to get f completely factored out as product of linear terms. Author said division could be long or synthetic. Up to here, it is pretty standard question on factor theorem. However, in solution 2 he just gave another way of writing f with each term containing (x-2) so that it can be factored out and got f(x) = (x-2) * quadratic and continue factoring quadratic to get f is product of linear terms.

 

[Frabjous]

It is given in the question that x=2 is a root, and solution 1 was straightforward by dividing f by (x-2) and factor resulting quadratic equation to get f completely factored out as product of linear terms. Author said division could be long or synthetic. Up to here, it is pretty standard question on factor theorem. However, in solution 2 he just gave another way of writing f with each term containing (x-2) so that it can be factored out and got f(x) = (x-2) * quadratic and continue factoring quadratic to get f is product of linear terms.

Unless there is more than the little you posted, what you are calling solution 2 is not a solution strategy. There are no steps, he just wrote down the answer.

 

[MiddleEast]

Exactly, thats my point it is at least non-standard way to simplify polynomial without standard long/synthetic vision method. This could help with higher level math questions when coe's are arbitrary constants.

 

[Frabjous]

Author said division could be long or synthetic.

I would guess that the second solution is the author doing synthetic long division in his/her head.

 

[MatinSAR]

Not deriving, main goal is rewriting polynomial in terms of (x-a) in each term in one single step so that (x-a) will be factored out.

Alright. Isn't that what I have done? I expressed the polynomial in terms of $x-2$ for each term.

rewriting polynomial in terms of (x-a) in each term in one single step so that (x-a) will be factored out.

You can factor $x-2$ out.

$2x^3-3x^2-17x+30=2x^3-4x^2+x^2-2x-15x+30$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ =2x^2(x-2)+x(x-2)-15(x-2)$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(x-2)(2x^2+x-15)$

 

[FactChecker]

It is given in the question that x=2 is a root,

It would have been nice to tell us that. I couldn't understand why you were so interested in dividing out x-2. In general, that does not help to answer your original question.

 

[DaveE]

Not sure if this is what you're looking for, but this is a power series about $x=2$.

Bonus points due for a method that isn't simple division. I take back what I said about that. I wouldn't have thought of this, I think.

OTOH, no one can do all those Taylor expansion derivatives in one step, or two.
$-5(x-2)+9(x-2)^2+2(x-2)^3$ isn't either the first or the last step here.

 

[symbolipoint]

Rational Roots Theorem! Several possible rational roots can be checked based on lead coefficient and the constant term. I avoid trying to describe here further.

 

[SammyS]

They are also skipping the most important step of determining that 2 is a rational root.

Fabulous point, @Frabjous !

It is given in the question that x=2 is a root, and solution 1 was straightforward by dividing f by (x-2) and factor resulting quadratic equation to get f completely factored out as product of linear terms. Author said division could be long or synthetic. Up to here, it is pretty standard question on factor theorem. However, in solution 2 he just gave another way of writing f with each term containing (x-2) so that it can be factored out and got f(x) = (x-2) * quadratic and continue factoring quadratic to get f is product of linear terms.

@MiddleEast
It would have been wonderful had you included this information in the OP.

 

[FactChecker]

It was obvious in the attached picture.

Things we find out from the solution should not be used to explain how to reach that solution. If the zero is not known ahead of time, it is not valid to use it in telling someone how to solve that problem. In addition, the special circumstance shown, where x=2 is a zero, is not valid for the general question asked: "How can you rewrite polynomial in terms of (x-a) instead of x?"