Remainder Theorem : Let \(p(x)\) be any polynomial of degree greater than or equal to one and let a be any real number. If \(p(x)\) is divided by the linear polynomial \(x-a\), then the remainder is \(p(a)\).
Proof : Let \(p(x)\) be any polynomial with degree greater than or equal to 1 . Suppose that when \(p(x)\) is divided by \(x-a\), the quotient is \(q(x)\) and the remainder is \(r(x)\), i.e.,
\[
p(x)=(x-a) q(x)+r(x)
\]
Since the degree of \(x-a\) is 1 and the degree of \(r(x)\) is less than the degree of \(x-a\), the degree of \(r(x)=0\). This means that \(r(x)\) is a constant, say \(r\).
So, for every value of \(x, r(x)=r\).
Therefore, \(\quad p(x)=(x-a) q(x)+r\)
In particular, if \(x=a\), this equation gives us
\[
\begin{aligned}
p(a) &=(a-a) q(a)+r \\
&=r,
\end{aligned}
\]
which proves the theorem.
Solution: Here, \(\quad p(x)=x^4+x^3-2 x^2+x+1\), and the zero of \(x-1\) is 1 .
So, \(\quad \begin{aligned} p(1) &=(1)^4+(1)^3-2(1)^2+1+1 \\ &=2 \end{aligned}\)
So, by the Remainder Theorem, 2 is the remainder when \(x^4+x^3-2 x^2+x+1\) is divided by \(x-1\)
Solution : As you know, \(q(t)\) will be a multiple of \(2 t+1\) only, if \(2 t+1\) divides \(q(t)\) leaving remainder zero. Now, taking \(2 t+1=0\), we have \(t=-\frac{1}{2}\).
Also, \(\quad q\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^3+4\left(-\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)-1=-\frac{1}{2}+1+\frac{1}{2}-1=0\) So the remainder obtained on dividing \(q(t)\) by \(2 t+1\) is 0 .
So, \(2 t+1\) is a factor of the given polynomial \(q(t)\), that is \(q(t)\) is a multiple of \(2 t+1\)
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