The polynomial remainder theorem may be used to evaluate {\displaystyle f(a)\,} f(a)\, by calculating the remainder, {\displaystyle r} r. Although polynomial long division is more difficult than evaluating the function itself, synthetic division is computationally easier. Thus, the function may be more "cheaply" evaluated using synthetic division and the polynomial remainder theorem.
The factor theorem is another application of the remainder theorem: if the remainder is zero, then the linear divisor is a factor. Repeated application of the factor theorem may be used to factorize the polynomial
In algebra, the polynomial remainder theorem is an application of Euclidean division of polynomials. It states that the remainder of the division of a polynomial {\displaystyle f(x)} f(x) by a linear polynomial {\displaystyle x-a} x-a is equal to {\displaystyle f(a).} f(a). In particular, {\displaystyle x-a} x-a is a divisor of {\displaystyle f(x)} f(x) if and only if {\displaystyle f(a)=0.}
Let f(x)=x^{3}-12x^{2}-42\,} f(x)=x^{3}-12x^{2}-42\,. Polynomial division of {\displaystyle f(x)\,} f(x)\, by x-3\,} x-3\, gives the quotient {2}-9x-27\,} x^{2}-9x-27\, and the remainder -123\,} -123\,. Therefore, {\displaystyle f(3)=-123\,} f(3)=-123\,.
Show that the polynomial remainder theorem holds for an arbitrary second degree polynomial f(x)=ax^{2}+bx+c} f(x)=ax^{2}+bx+c by using algebraic manipulation:
{f(x)}{x-r}}&={\frac {a{x^{2}}+bx+c}{x-r}}\\&={\frac {a{x^{2}}-arx+arx+bx+c}{x-r}}\\&={\frac {ax(x-r)+(b+ar)x+c}{x-r}}\\&=ax+{\frac {(b+ar)(x-r)+c+r(b+ar)}{x-r}}\\&=ax+b+ar+{\frac {c+r(b+ar)}{x-r}}\\&=ax+b+ar+{\frac {a{r^{2}}+br+c}{x-r}}\end{aligned}}} {\begin{aligned}{\frac {f(x)}{{x-r}}}&={\frac {{a{x^{2}}+bx+c}}{{x-r}}}\\&={\frac {{a{x^{2}}-arx+arx+bx+c}}{{x-r}}}\\&={\frac {{ax(x-r)+(b+ar)x+c}}{{x-r}}}\\&=ax+{\frac {{(b+ar)(x-r)+c+r(b+ar)}}{{x-r}}}\\&=ax+b+ar+{\frac {{c+r(b+ar)}}{{x-r}}}\\&=ax+b+ar+{\frac {{a{r^{2}}+br+c}}{{x-r}}}\end{aligned}}
Multiplying both sides by (x − r) gives
{\displaystyle f(x)=ax^{2}+bx+c=(ax+b+ar)(x-r)+{a{r^{2}}+br+c}} f(x)=ax^{2}+bx+c=(ax+b+ar)(x-r)+{a{r^{2}}+br+c}.
Since {\displaystyle R=ar^{2}+br+c} R=ar^{2}+br+c is our remainder, we have indeed shown that {\displaystyle f(r)=R} f(r)=R.
The Remainder Theorem is useful for evaluating polynomials at a given value of x, though it might not seem so, at least at first blush. This is because the tool is presented as a theorem with a proof, and you probably don't feel ready for proofs at this stage in your studies. Fortunately, you don't "have" to understand the proof of the Theorem; you just need to understand how to use the Theorem.
The Remainder Theorem starts with an unnamed polynomial p(x), where "p(x)" just means "some polynomial p whose variable is x". Then the Theorem talks about dividing that polynomial by some linear factor x – a, where a is just some number. Then, as a result of the long polynomial division, you end up with some polynomial answer q(x) (the "q" standing for "the quotient polynomial") and some polynomial remainder r(x).
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The Remainder Theorem
The Remainder Theorem is useful for evaluating polynomials at a given value of x, though it might not seem so, at least at first blush. This is because the tool is presented as a theorem with a proof, and you probably don't feel ready for proofs at this stage in your studies. Fortunately, you don't "have" to understand the proof of the Theorem; you just need to understand how to use the Theorem.
The Remainder Theorem starts with an unnamed polynomial p(x), where "p(x)" just means "some polynomial p whose variable is x". Then the Theorem talks about dividing that polynomial by some linear factor x – a, where a is just some number. Then, as a result of the long polynomial division, you end up with some polynomial answer q(x) (the "q" standing for "the quotient polynomial") and some polynomial remainder r(x).
As a concrete example of p, a, q, and r, let's look at the polynomial p(x) = x3 – 7x – 6, and let's divide by the linear factor x – 4 (so a = 4):
completed division: quotient x^2 + 4x + 9, remainder 30
So we get a quotient of q(x) = x2 + 4x + 9 on top, with a remainder of r(x) = 30.
You know, from long division of regular numbers, that your remainder (if there is one) has to be smaller than whatever you divided by. In polynomial terms, since we're dividing by a linear factor (that is, a factor in which the degree on x is just an understood "1"), then the remainder must be a constant value. That is, when you divide by "x – a", your remainder will just be some number.
The Remainder Theorem then points out the connection between division and multiplication. For instance, since 12 ÷ 3 = 4, then 4 × 3.
