Remainder/factor theorem question

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To find the values of "a" and "b" in the polynomial f(x) = ax^3 - bx^2 + 2x - 12, given the divisor x^2 - 5x + 6 and the remainder 2x - 3, polynomial long division can be used effectively. By setting up the equation f(x) = (cx + d)(x^2 - 5x + 6) + (2x - 3), one can manipulate the equation to isolate the unknowns. Subtracting the remainder from f(x) allows for easier application of the division statement, leading to the determination of "a" and "b." The divisor is factorable, which can further simplify the process using the factor theorem. This method provides a structured approach to solving for the unknown coefficients in the polynomial.
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The question is
Find "a" and "b":

f(x) is
ax^3 - bx^2 + 2x - 12

Divisor is
x^2 - 5x +6

Remainder is
2x - 3So what I tried to do was divide f(x) with the divisor and take the the remainder that I got and let it = 2x - 3 but I always get stuck in the end. My teacher suggested using the division statement but I'm not quite sure how that will help.

Thanks for helping!
 
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One method to solve this is use polynomial long division starting at the end, where you know the remainder, the divisor, and the last two terms of the dividend, which allows you to determine the last term of the quotient, then work backwards until you solve for b then a.
 
ouu okay I see!
But is there a shorter method to this?

Thanks in advance
 
What you have is

ax^3 - bx^2 + 2x - 12 = (cx + d) (x^2 - 5x + 6) + (2x - 3)

where a, b, c and d are unknown. You could subract (2x - 3) from both sides first, then use the long division method or inspection to determine the values, but subtracting (2x - 3) from both sides is the same as what you're doing in the first step of working the long division backwards. This backwards long division method is used in general for "division problems" when there are unknowns in the dividend, divisor, and/or remainder.
 
Last edited:
HerroFish said:
The question is
Find "a" and "b":

f(x) is
ax^3 - bx^2 + 2x - 12

Divisor is
x^2 - 5x +6

Remainder is
2x - 3

So what I tried to do was divide f(x) with the divisor and take the the remainder that I got and let it = 2x - 3 but I always get stuck in the end. My teacher suggested using the division statement but I'm not quite sure how that will help.

Thanks for helping!
So, If f(x) divided by g(x) yields a quotient q(x) and remainder r(x), then

f(x)=g(x)\cdot q(x)+r(x)\ .

Therefore, f(x)-r(x)=g(x)\cdot q(x)\ . So that g(x) divides (f(x)-r(x)) with a remainder of zero.

Notice that in your problem the divisor, x2 - 5x +6, is factorable. Apply the factor theorem to f(x)-r(x), with each of those factors.
 
:biggrin:
SammyS said:
So, If f(x) divided by g(x) yields a quotient q(x) and remainder r(x), then

f(x)=g(x)\cdot q(x)+r(x)\ .

Therefore, f(x)-r(x)=g(x)\cdot q(x)\ . So that g(x) divides (f(x)-r(x)) with a remainder of zero.

Notice that in your problem the divisor, x2 - 5x +6, is factorable. Apply the factor theorem to f(x)-r(x), with each of those factors.

Okay I get it now!
Thanks everybody for all your help! :smile: :!)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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