Factot Theorem

[loz58]

How do you factorise
this polynomial using the factor theorem

(x-2) is the factor of 2x3 +x2 -13x +6

Thanks

[gen2]

consider solving the equation
2x^3 + x^2 -13x +6 = 0

If (x - 2) is a factor of the above, then it can be written
(x - 2) (f(x)) = 0
so either (x -2) = 0 or f(x) = 0

If x - 2 = 0 then x = 2

Put x = 2 into the original equation and see if it equals 0.
If it does, then (x - 2) was a factor of the given expression.

[loz58]

thank you very much for you help, but do you know now how to factorise the (x-2) to find the two other possible factors for
2x3 + x2 - 13x + 6

[gen2]

OK. So you have 2x^3 + x^2 -13x +6 and you now know (x-2) is a factor so your next stage is to determine the other factors.

So
(x - 2) (ax^2 + bx + c) = 0

Since this has to expand to 2x^2 + . . . etc
then x multiplied by ax^2 has to = 2x^3
so therefore a = 2

(x - 2) (2x^2 + bx + c) = 0

Now to determine b, look at the x^2 term
(x) multipied by (bx) + (-2) times (2x^2) has to = + x^2
so bx^2 -4x^2 = x^2
so b = 5

(x - 2) (2x^2 + 5x + c) = 0

Now to determine c, look at the term without x

(-2) multipied by (c) = 6
so c = -3

That gives (x - 2) (2x^2 + 5x - 3) = 0
Double-check by expanding this to confirm that all terms agree.
. . . . . . . . . . . . . . . . . . . .

(x - 2) (2x^2 + 5x - 3) = 0
so either (x - 2) = 0 . . . so x = 2 is one solution
or (2x^2 + 5x - 3) = 0

You could now attempt to factorise the quadratic (2x^2 + 5x - 3) but there are no integer factors.

You will have to use the formula for solving a quadratic to give the remaining two roots

x = (-b +/- SQR (b^2 - 4ac)) / 2a

This wil give you the remaining two solutions.</b

[dundurn]

I agree there are no integer soutions but the factorisation is still reasonably straightforward (2x-1)(x+3) so taking the whole original expression (x-2)(2x-1)(x+3) the roots of the equation are x=2, x=0.5 and x=-3

[loz58]

thank you very much!
that has helped me so much
thanks again

[Peter Pedant]

You can also do a long diviison of 2x3+x2-13x+6 of x-2

and you do it in the classic fashion:

x+2 into 2x3 goes 2x2 times, and x+2 times 2x2 is 2x3+4x2 and so you write:


_ 2x2__________
x-2 I 2x3 + x2 -13x + 6
2x3 - 4x2
--------------
-5x2 this is where you subtract

and then you say x-2 into -5x2 goes -5x times and -5x times x-2 is -5x2 + 10
put that under and subtract....
and I think along the top you will generate

2x2 - 5x +3 along the top