If (2x + 1) is a factor of both the expressions 2x² - 5x + p and 2x² + 5x + q, find the values of p and q. Hence, find the other factors of both the polynomials.

By factor theorem (x - b) is a factor of f(x), if f(b) = 0.

Let, f(x) = 2x² - 5x + p

Given, (2x + 1) or 2(x - (-$\dfrac{1}{2}$) is a factor of f(x)

$\therefore f(-\dfrac{1}{2}) = 0 \\[1em] \Rightarrow 2\big(-\dfrac{1}{2}\big)^2 - 5\big(-\dfrac{1}{2}\big) + p = 0 \\[1em] \Rightarrow 2\big(\dfrac{1}{4}) + \dfrac{5}{2} + p = 0 \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{5}{2} + p = 0 \\[1em] \Rightarrow \dfrac{6}{2} + p = 0 \\[1em] \Rightarrow 3 + p = 0 \\[1em] p = -3$

Putting value of p in f(x)

$f(x) = 2x^2 - 5x - 3 \\[0.5em] = 2x^2 - 6x + x - 3 \\[0.5em] = 2x(x - 3) + 1(x - 3) \\[0.5em] = (2x + 1)(x - 3)$

Hence, p = -3 and other factor is (x - 3).

Let, g(x) = 2x² + 5x + q

Given, (2x + 1) or 2(x - (-$\dfrac{1}{2}$) is a factor of g(x)

$\therefore g(-\dfrac{1}{2}) = 0 \\[0.5em] \Rightarrow 2\big(-\dfrac{1}{2}\big)^2 + 5\big(-\dfrac{1}{2}\big) + q = 0 \\[0.5em] \Rightarrow 2\big(\dfrac{1}{4}) - \dfrac{5}{2} + q = 0 \\[0.5em] \Rightarrow \dfrac{1}{2} - \dfrac{5}{2} + q = 0 \\[0.5em] \Rightarrow -\dfrac{4}{2} + q = 0 \\[0.5em] \Rightarrow -2 + q = 0 \\[0.5em] q = 2$

Putting value of q in g(x)

$f(x) = 2x^2 + 5x + 2 \\[0.5em] = 2x^2 + 4x + x + 2 \\[0.5em] = 2x(x + 2) + 1(x + 2) \\[0.5em] = (2x + 1)(x + 2)$

Hence, q = 2 and other factor is (x + 2).