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.

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.

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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