====== Question 3 and 4, Exercise 5.2 ======
Solutions of Question 3 and 4 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 3=====
Factorize by using factor theorem: $2 x^{3}+5 x^{2}-9 x-18$

** Solution. **

Suppose \( f(x) = 2x^{3} + 5x^{2} - 9x - 18 \).

\begin{align*}
f(-2) &= 2(-2)^{3} + 5(-2)^{2} - 9(-2) - 18 \\
&= 2(-8) + 5(4) + 18 - 18 \\
&= -16 + 20 + 18 - 18 = 0.
\end{align*}

By the factor theorem, \( x + 2 \) is a factor of \( f(x) \).

Using synthetic division:
\[
\begin{array}{r|rrrr}
-2 & 2 & 5 & -9 & -18 \\
&    & -4 & -2 & 22 \\
\hline
& 2  & 1  & -11 & 0 \\
\end{array}
\]

This gives:
\begin{align*}
f(x) &= (x + 2)(2x^{2} + x - 9).
\end{align*}
Thus, we can factor \( 2x^{2} + x - 9 \) as:
\begin{align*}
(x + 2)(2x^{2} + x - 9) &=(x + 2)( 2x^{2} + 6x - 3x - 9) \\
&= (x + 2)[2x(x + 3) - 3(x + 3)] \\
&= (x + 2)(2x - 3)(x + 3).
\end{align*}
The solution set is $$f(x)=(x + 2)(2x - 3)(x + 3).$$

=====Question 4=====
Factorize by using factor theorem: $3 x^{3}-5 x^{2}-36$

**Solution.**

Suppose \( f(x) = 3x^{3} - 5x^{2} - 36 \).

\begin{align*}
f(3) &= 3(3)^{3} - 5(3)^{2} - 36 \\
 &= 3(27) - 5(9) - 36 \\
 &= 81 - 45 - 36 = 0.
\end{align*}

By the factor theorem, \( x - 3 \) is a factor of \( f(x) \).

Using synthetic division:
\[
\begin{array}{r|rrrr}
3 & 3 & -5 & 0 & -36 \\
 &    & 9  & 12 & 36  \\
\hline
& 3  & 4  & 12 & 0 \\
\end{array}
\]

This gives:
\begin{align*}
f(x) &= (x - 3)(3x^{2} + 4x + 12).
\end{align*}
The final factorization is:
\begin{align*}
f(x) &= (x - 3)(3x^{2} + 4x + 12).
\end{align*}










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