Question 8 and 9, Exercise 5.1
Solutions of Question 8 and 9 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 8
Find zeros of the polynomial $2 x^{3}+3 x^{2}-11 x-6$.
Solution.
Suppose $p(x)=2x^3+3x^2-11x-6$.
Since
\begin{align}
p(2) &= 2(2)^3+3(2)^2-11(2)-6 \\
&=16+12-22-6 = 0 \end{align}
Hence 2 is zero of $p(x)$.
Then by using synthetic division:
\begin{align}
\begin{array}{r|rrrr}
2 & 2 & 3 & -11 & -6 \\
& \downarrow & 4 & 14 & 6 \\
\hline
& 2 & 7 & 3 & 0 \\
\end{array}\end{align}
Now
\begin{align*}
& 2x^2+7x+3 \\
= & 2x^2+6x+x+3 \\
= & 2x(x+3)+1(x+3) \\
= & (x+3)(2x+1)
\end{align*}
i.e. $x+3=0$ or $2x+1=0$
$\implies$ $x=-3$ or $x=-\dfrac{1}{2}$.
Hence $2$, $3$ and $-\dfrac{1}{2}$ are the roots of given polynomial.
Question 9
Express $f(x)=x^{3}-x^{2}-14 x+11$ in the form $f(x)=(x-a) q(x)+r$, where $a=4$. ( statement corrected).
Solution.
By synthetic division \begin{align} \begin{array}{r|rrrr} 4 & 1 & -1 & -14 & 11 \\ & \downarrow & 4 & 12 & -8 \\ \hline & 1 & 3 & -2 & |\,\, 3 \\ \end{array}\end{align}
Hence $$f(x)=(x-2)(x^2+3x-2)+3$$
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