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Question 5 and 6, Exercise 5.2
Solutions of Question 5 and 6 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.
Question 5
Factorize by using factor theorem: $t^{3}+t^{2}+3 t-5$
Solution.
Suppose \( f(t) = t^{3} + t^{2} + 3t - 5 \).
\begin{align*} f(1) &= (1)^{3} + (1)^{2} + 3(1) - 5 \\ &= 1 + 1 + 3 - 5 \\ &= 0. \end{align*}
By the factor theorem, \( t - 1 \) is a factor of \( f(t) \).
Using synthetic division:
\begin{align} \begin{array}{r|rrrr} 1 & 1 & 1 & 3 & -5 \\ & & 1 & 2 & 5 \\ \hline & 1 & 2 & 5 & 0 \\ \end{array} \end{align}
This gives: \begin{align*} f(t) &= (t - 1)(t^{2} + 2t + 5). \end{align*} Thus, the factorization is: \begin{align*} f(t) &= (t - 1)(t^{2} + 2t + 5). \end{align*}
Question 6
If $(x-2)$ is one of the factor of $2 x^{3}-15 x^{2}+16 x+12$, find its other factors.
Solution.
It is given by the factor theorem, \( x - 2 \) is a factor of \( f(x) \).Then
Using synthetic division to divide \( f(x) \) by \( x - 2 \):
\begin{align} \begin{array}{r|rrrr} 2 & 2 & -15 & 16 & 12 \\ & & 4 & -22 & -12 \\ \hline & 2 & -11 & -6 & 0 \\ \end{array} \end{align}
This gives: \begin{align*} f(x) &= (x - 2)(2x^{2} - 11x - 6). \end{align*} \begin{align*} 2x^{2} - 11x - 6 &= 2x^{2} - 12x + x - 6 \\ &= 2x(x - 6) + 1(x - 6) \\ &= (2x + 1)(x - 6). \end{align*}
Finally, the complete factorization is: \begin{align*} f(x) &= (x - 2)(2x + 1)(x - 6). \end{align*}
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