Question

If g(x) = (2x – 3) is a factor of f(x) = 2x³ – 9x2 + x + p, find the value of p. Hence factirize f(x).

Answer 1

Solution:

If g(x) is a factor of f(x),than it satisfies f(x)

So,here I am going to apply remainder theorem,in which we put the value of x from g(x) into f(x)

$g(x) = 0 \\ \\ 2x - 3 = 0 \\ \\ 2x = 3 \\ \\ x = \frac{3}{2}$
Now put this value in f(x)

$2 {x}^{3} - 9 {x}^{2} + x + p \\ \\ 2 ({ \frac{3}{2} })^{3} - 9( { \frac{3}{2} )}^{2} + ( \frac{3}{2} ) + p = 0 \\ \\ \frac{27}{4} - \frac{81}{4} + \frac{3}{2} + p = 0 \\ \\ p = \frac{81}{4} - \frac{27}{4} - \frac{3}{2} \\ \\ p= \frac{81- 27 - 6}{4} \\ \\ p = \frac{48}{4} \\ \\ p = 12$
Now to factorise f(x)

$2 {x}^{3} - 9 {x}^{2} + x + 12 \\ \\2x - 3) 2{x}^{3} - 9{x}^{2} + x + 12( {x}^{2} - 3x - 4 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: 2 {x}^{3} - 3 {x}^{2} \\ - - - - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 6 {x}^{2} + x \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: - 6 {x}^{2} + 9x \\ \: \: \: \: \: \: \: \: \: \: \: \: - - - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: - 8x + 12 \\ \: \: \: \: \: \: \: \: \: \: \: \: - 8x + 12 \\ \: \: \: \: - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \:$
Now factorise

${x}^{2} - 3x - 4 \\ \\ {x}^{2} - 4x + x - 4 \\ \\ x(x - 4) + 1(x - 4) \\ \\ (x - 4)(x + 1)$
So factors of f(x) are
$(2x - 3)(x - 4)(x + 1)$
Hope it helps you