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remainder theorem

f(𝑥)=2𝑥3−6𝑥2−2𝑥+6


(x-3) is a factor

Q) factorise f(x) fully.

I have no idea how to do this....
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TenOfThem
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Original post by 50ShadesOfRay



(x-3) is a factor

Q) factorise f(x) fully.

I have no idea how to do this...


polynomial division is one option
(edited 10 years ago)
I dont think it is remainder theorem question; you can use remainder theorem to prove (x-3) is factor by using f(3); if you get answer of zero when substituting x for 3, then its a factor.
After that use long division to divide f(x) by (x-3) to get a quadratic quotient. Then use quadratic formula to factorise quadratic quotient...
Original post by 50ShadesOfRay
f(𝑥)=2𝑥3−6𝑥2−2𝑥+6


(x-3) is a factor

Q) factorise f(x) fully.

I have no idea how to do this....


Use synthetic division, it saves a lot of time :wink:

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Original post by 50ShadesOfRay
f(𝑥)=2𝑥3−6𝑥2−2𝑥+6


(x-3) is a factor

Q) factorise f(x) fully.

I have no idea how to do this....


You should realise that the other factor must be a quadratic so assume it is (ax2+bx+c)(ax^2+bx+c)
You now want (x−3)(ax2+bx+c)=2x3−6x2−2x+6(x-3)(ax^2+bx+c)=2x^3-6x^2-2x+6
Now simply compare the coefficients of the different powers of x.
For example, multiplying out the two brackets we will get ax3 ax^3 so we must have a=2. Repeat for the coefficients of x2 and xx^2 \mathrm{\ and\ }x to get b and c.
Original post by majmuh24
Use synthetic division, it saves a lot of time :wink:

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I agree it's amazing how much time it saves! :eek:
Original post by ThePerfectScore
I agree it's amazing how much time it saves! :eek:


Yeah, I remember when I used to equate coefficients and use polynomial long division :moon: Not any more :tongue:

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