Factorising a cubic polynomial

[PlasmaSphere]

1. A cubic polynomial is given by (f)x = x³ + x² - 10x + 8

1) i) Show that (x -1) is a factor of f(x)

ii) Factorise f(x) fully

iii) Sketch the graph



2. Factor theorem?



3. Attempt at solution

i) I know that f(1) = 0, so (x - 1) is a factor of (f)x = x³ + x² - 10x + 8

ii) I think i am correct in saying that since you know (x - 1) is a factor you can write it like this; ( x-1 )( x² - 10x + 8 ), i just don't know where to go from there. I don't think you can factorise x² - 10x + 8 any further?

I presume that by factorising fully, they mean in the form (x-a)(x-b)(x-c), i just can't seem to get there.

 

[stunner5000pt]

i) is correct

ii) is not because you cannot just jump to that answer. To get the factored form, you must divide f(x) by x-1 either by long division or synthetic division and then get the other factor.

 

[HallsofIvy]

1. A cubic polynomial is given by (f)x = x³ + x² - 10x + 8

1) i) Show that (x -1) is a factor of f(x)

ii) Factorise f(x) fully

iii) Sketch the graph



2. Factor theorem?



3. Attempt at solution

i) I know that f(1) = 0, so (x - 1) is a factor of (f)x = x³ + x² - 10x + 8

ii) I think i am correct in saying that since you know (x - 1) is a factor you can write it like this; ( x-1 )( x² - 10x + 8 ), i just don't know where to go from there. I don't think you can factorise x² - 10x + 8 any further?

How did you get that? Just by dropping the x³? Now you know that's not right! As Plasmasphere said, you could divide $x^ x^2- 10x+ 8$ by x-1. Or you can "work backwards. You want to find $ax^2+ bx+ c$ so that $(x-1)(ax^2+ bx+ c)= x^3+ x^2- 10x+ 8$. Obviously, since $x(x^2)= x^3$ and $-1(-8)= 8$, a= 1, c= -8 (NOT +8!). Okay, multiply out $(x-1)(x^2+ bx- 8)= x^3+ x^2- 20x+ 8$ to see what b must be.

I presume that by factorising fully, they mean in the form (x-a)(x-b)(x-c), i just can't seem to get there.

 

[kbaumen]

Take a look https://www.physicsforums.com/showthread.php?t=196849".

 

[silver-rose]

http://en.wikipedia.org/wiki/Polynomial_long_division

Is the wiki page for the technique used in the link given by kbaumen.

You should divide the original polynomial by the given factor; that would leave you with a quadratic polynomial, which you should be able to factor without any difficulty.

 

[PlasmaSphere]

Thanks for all the help, i used the working backwards method, and i got the right answer according to the answer booklet, the sketch for part iii) was easy too.

I'm not sure why i thought that i could write it as; ( x-1 )( x² - 10x + 8 ), if you expand that it clearly will never give the correct answer. I was jumping two steps ahead.

Cheers, those pages on polynomial long division should be helpful in the future too.

 

[The_ArtofScience]

Okay, multiply out $(x-1)(x^2+ bx- 8)= x^3+ x^2- 20x+ 8$ to see what b must be.

I don't understand that step (?) just curious how you got there

 

[HallsofIvy]

If you are wondering about the "20x", that was a typo. The right hand side should be just the original polynomial, $x^3+ x^2-10x+ 8$.

 

[kbaumen]

http://en.wikipedia.org/wiki/Polynomial_long_division

Is the wiki page for the technique used in the link given by kbaumen.

You should divide the original polynomial by the given factor; that would leave you with a quadratic polynomial, which you should be able to factor without any difficulty.

Actually I meant the Horner's scheme to be used for factorizing the equation. Sorry, if I made that unclear.

 

[Kenny09]

What if we take division of the polynomial ?
( x3 + x2 - 10x + 8 ) / ( x-1)

 

[HallsofIvy]

What if we take division of the polynomial ?
( x3 + x2 - 10x + 8 ) / ( x-1)

Yes, that's already been discussed in several responses.