While factorizing a given polynomial, using remainder and factor theorem, a student finds that (2x + 1) is a factor of 2x3 + 7x2 + 2x - 3.
(a) Is the student's solution correct stating that (2x + 1) is a factor of the given polynomial ?
(b) Given a valid reason for your answer.
Also, factorize the given polynomial completely.
Show SolutionHide Solution↓
⇒ 2x + 1 = 0
⇒ 2x = -1
⇒ x = $-\frac{1}{2}$
Substituting x = $-\frac{1}{2}$ in 2x3 + 7x2 + 2x - 3, we get :
$$\begin{aligned}\Rightarrow 2 \times \Big(-\frac{1}{2}\Big)^3 + 7 \times \Big(-\frac{1}{2}\Big)^2 + 2 \times \Big(-\frac{1}{2}\Big) - 3 \\[1em] \Rightarrow 2 \times -\frac{1}{8} + 7 \times \frac{1}{4} + (-1) - 3 \\[1em] \Rightarrow -\frac{1}{4} + \frac{7}{4} - 4 \\[1em] \Rightarrow \frac{-1 + 7}{4} - 4 \\[1em] \Rightarrow \frac{6}{4} - 4 \\[1em] \Rightarrow \frac{6 - 16}{4} \\[1em] \Rightarrow \frac{-10}{4} \\[1em] \Rightarrow -\frac{5}{2}.\end{aligned}$$
Since, remainder is not equal to zero.
Hence, (2x + 1) is not a factor of the given polynomial.
Substituting x = $\frac{1}{2}$ in 2x3 + 7x2 + 2x - 3, we get :
$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{1}{2}\Big)^3 + 7 \times \Big(\frac{1}{2}\Big)^2 + 2 \times \frac{1}{2} - 3 \\[1em] \Rightarrow 2 \times \frac{1}{8} + 7 \times \frac{1}{4} + 1 - 3 \\[1em] \Rightarrow \frac{1}{4} + \frac{7}{4} - 2 \\[1em] \Rightarrow \frac{8}{4} - 2 \\[1em] \Rightarrow 2 - 2 \\[1em] \Rightarrow 0.\end{aligned}$$
Since, remainder is equal to zero.
∴ x - $\frac{1}{2}$ is factor of polynomial,
⇒ x - $\frac{1}{2}$ = 0
⇒ x = $\frac{1}{2}$
⇒ 2x = 1
⇒ 2x - 1 is factor of polynomial.
Dividing 2x3 + 7x2 + 2x - 3 by 2x - 1, we get :
$$\begin{array}{l} \phantom{2x - 1)}{\quad x^2 + 4x + 3} \\ 2x - 1\overline{\smash{\big)}\quad 2x^3 + 7x^2 + 2x - 3} \\ \phantom{2x - 1)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \\ \phantom{{2x - 1}+2x^3 + 1}8x^2 + 2x \\ \phantom{{2x - 1)}+2x^31}\underline{\underset{-}{+}8x^2 \underset{+}{-} 4x} \\ \phantom{{2x - 1)}+2x^31+8x^212}6x - 3 \\ \phantom{{2x - 1)}+2x^31+8x^21}\underline{\underset{-}{+}6x \underset{+}{-} 3} \\ \phantom{{2x - 1)}+2x^31+8x^21+6}\times \end{array}$$
2x3 + 7x2 + 2x - 3 by 2x - 1 = (2x - 1)(x2 + 4x + 3)
= (2x - 1)[x2 + 3x + x + 3]
= (2x - 1)[x(x + 3) + 1(x + 3)]
= (2x - 1)(x + 1)(x + 3).
Hence, 2x3 + 7x2 + 2x - 3 = (2x - 1)(x + 1)(x + 3).