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.

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 $$f(x)=(x + 2)(2x - 3)(x + 3).$$

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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