Question

If ax³+bx²+x-3 has x+2 as a factor and leaves remainder 4when divided by x-2 find values of a and b

Answer 1

GIVEN :–

• A Polynomial P(x) = ax³ + bx² + x - 3 have a factor (x + 2).

• When polynomial divided by (x - 2) and leave 4 as remainder.

TO FIND :–

• Value of 'a' and 'b' = ?

SOLUTION :–

▪︎According to the first condition –

• If (x + 2) is a factor of Polynomial , then x = 2 will satisfy the equation.

$\\ \implies \sf p( -2 ) = 0$

$\\ \implies \sf a {( - 2)}^{3} + b {( - 2)}^{2} - 2 - 3 = 0$

$\\ \implies \sf - 8a + 4b - 5 = 0$

$\\ \implies \sf - 8a + 4b = 5$

$\\ \implies \sf 4b = 5 + 8a$

$\\ \implies \sf b = \dfrac{1}{4} (5 + 8a) \: \: \: - - - eq.(1)$

▪︎According to the second condition –

$\\ \implies \sf p(2) = 4$

$\\ \implies \sf a {( 2)}^{3} + b {(2)}^{2} + 2 - 3 = 0$

$\\ \implies \sf 8a+4b= 1$

• Using eq.(1) –

$\\ \implies \sf 8a+4 \left[ \dfrac{1}{4}(5 + 8a) \right] = 1$

$\\ \implies \sf 8a+(5 + 8a) = 1$

$\\ \implies \sf 16a = 1 - 5$

$\\ \implies \sf 16a = - 4$

$\\ \implies \sf a = \dfrac{ - 4}{16}$

$\\ \implies \large { \boxed{ \sf a = - \dfrac{1}{4}}}$

• Put the value of 'a' in eq.(1) –

$\\ \implies \sf b = \dfrac{1}{4} \left(5 + 8 (- \frac{1}{4}) \right)$

$\\ \implies \sf b = \dfrac{1}{4} \left(5 - 2 \right)$

$\\ \implies \sf b = \dfrac{1}{4} \left(3 \right)$

$\\ \implies \large { \boxed{ \sf b= \dfrac{3}{4}}}$