====== Question 6 and 7, Exercise 5.1 ======
Solutions of Question 6 and 7 of Exercise 5.1 of Unit 05: Polynomials. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 6=====
Find the value of ' $m$ ' in the polynomial $2 x^{3}+3 x^{2}-3 x-m$ which when divided by $x-2$ gives the remainder of 16 .

** Solution. **

Let $p(x)=2 x^{3}+3 x^{2}-3 x-m$ and $x-c=x-2$ $\implies c=2$. \\
By the Remainder Theorem, we have
\begin{align*}
\text{Remainder} & = p(c) = p(2) \\
& = 2(2)^{3} + 3(2)^{2} - 3(2) - m \\
& = 2(8) + 3(4) - 3(2) - m \\
& = 16 + 12 - 6 - m \\
& = 22 - m.
\end{align*}
Given that the remainder is 16, so
\begin{align*}
22 - m & = 16 \\
m & = 22 - 16 \\
m & = 6.
\end{align*}
Hence, the value of \( m \) is 6. GOOD





=====Question 7=====
Check whether 1 and -2 are the zeros of $x^{3}-7 x+6$.

** Solution. **
Suppose $p(x)=x^3-7x+6$.\\
$1$ will be zero of $p(x)$ if $p(1)=0$. Thus
\begin{align*}
p(1)&=(1)^3-7(1)+6\\
&=1-7+6\\
&=0\end{align*}
Hence $1$ is the zero of p(x). \\
Similarly, $-2$ will be zero of $p(x)$ if f $p(-2)=0$.
\begin{align*}
p(-2)&=(-2)^3-7(-2)+6\\
&=-8+14+6\\
&=12 \neq 0\end{align*}
This gives -2 is not zero of $p(x)$.\\
Hence only $1$ is the zero of P(x). GOOD









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