Given f(x) = ax² + bx + 2 and g(x) = bx² + ax + 1. If (x - 2) is a factor of f(x) but leaves the remainder -15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression

f(x) + g(x) + 4x² + 7x

f(x) = ax² + bx + 2

Given, (x - 2) is a factor of f(x) hence, by factor theorem f(2) = 0

$\therefore a(2)^2 + b(2) + 2 = 0 \\[0.5em] \Rightarrow 4a + 2b + 2 = 0$

On dividing equation by 2,

$\Rightarrow 2a + b + 1 = 0 \\[0.5em] b = -1 - 2a \text{( Equation 1)}$

g(x) = bx² + ax + 1

Given, on dividing g(x) by (x - 2), remainder = -15 and by remainder theorem, remainder = g(2)

$\therefore g(2) = -15 \\[0.5em] \Rightarrow b(2)^2 + a(2) + 1 = -15 \\[0.5em] \Rightarrow 4b + 2a = -15 - 1 \\[0.5em] \Rightarrow 4b + 2a = -16$

On dividing equation by 2,

$\Rightarrow 2b + a = -8 \\[0.5em]$

Putting value of b = -1 - 2a from equation 1,

$\Rightarrow 2(-1 - 2a) + a = -8 \\[0.5em] \Rightarrow -2 - 4a + a = -8 \\[0.5em] \Rightarrow -3a = -8 + 2 \\[0.5em] \Rightarrow a = \dfrac{-6}{-3} \\[0.5em] \Rightarrow a = 2 \\[0.5em] \text{ and } b = -1 - 2a = -1 - 4 = -5.$

Putting value of a = 2 and b = -5 in f(x) + g(x) + 4x² + 7x we get,

$(2x^2 - 5x + 2) + (-5x^2 + 2x + 1) + 4x^2 + 7x \\[0.5em] = 2x^2 - 5x^2 + 4x^2 - 5x + 2x + 7x + 2 + 1 \\[0.5em] = x^2 + 4x + 3 \\[0.5em] = x^2 + 3x + x + 3 \\[0.5em] = x(x + 3) + 1(x + 3) \\[0.5em] = (x + 1)(x + 3)$

Hence, the value of a = 2 and b = -5; f(x) + g(x) + 4x² + 7x = (x + 1)(x + 3).