Question 1 and 2, Exercise 5.2

Solutions of Question 1 and 2 of Exercise 5.2 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.

Factorize by using factor theorem: $y^{3}-7 y-6$

Solution.

Suppose $f(y)=y^{3}-7 y-6$.

\begin{align*} f(-1)&=(-1)^{3}-7 (-1)-6 \\ &= -1+7-6 =0. \end{align*} By factor theorem, $y+1$ is factor of $f(y)$.

Using synthetic division: \begin{align} \begin{array}{r|rrrr} -1 & 1 & 0 & -7 & -6 \\ & \downarrow & -1 & 1 & 6 \\ \hline & 1 & -1 & -6 & 0 \\ \end{array}\end{align}

This gives \begin{align*} f(y)& =(y+1)(y^2-y-6) \\ & = (y+1)(y^2-3y+2y-6) \\ & = (y+1)(y(y-3)+2(y-3)) \\ & = (y+1)(y-3)(y+2). \end{align*}

Factorize by using factor theorem: $2 x^{3}-x^{2}-2 x+1$

Solution.

\begin{align*} f(x) &= 2x^{3} - x^{2} - 2x + 1 \\ f(1) &= 2(1)^{3} - (1)^{2} - 2(1) + 1 \\ &= 2 - 1 - 2 + 1 = 0. \end{align*}

By the factor theorem, \( x - 1 \) is a factor of \( f(x) \).

Using synthetic division: \[ \begin{array}{r|rrrr} 1 & 2 & -1 & -2 & 1 \\ & & 2 & 1 & -1 \\ \hline & 2 & 1 & -1 & 0 \\ \end{array} \]

\begin{align*} f(x) &= (x - 1)(2x^{2} + x - 1) \\ 2x^{2} + x - 1 &= 2x^{2} + 2x - x - 1 \\ &= 2x(x + 1) - 1(x + 1) \\ &= (2x - 1)(x + 1). \end{align*}

Thus, the complete factorization is: \[ f(x) = (x - 1)(2x - 1)(x + 1). \]

Question 3 & 4>