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.

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.

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).

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