;====== 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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