The expression $2 x^{3} + b x^{2} - c x + d$ leaves the same remainder, when divided by $x + 1$ or $x - 2$ or $2 x - 1$ . Find $b$ and $c$ .

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Solution

$(x + 1) (x - 2) (2 x - 1) = 2 x^{3} + b x^{2} - c x + d$
$\Rightarrow (x^{2} - 2 x + x - 2) (2 x - 1) = 2 x^{3} + b x^{2} - c x + d$
$\Rightarrow (x^{2} - x - 2) (2 x - 1) = 2 x^{3} + b x^{2} - c x + d$
$\Rightarrow 2 x^{3} - x^{2} - 2 x^{2} + x - 4 x + 2 = 2 x^{3} + b x^{2} - c x + d$
$\Rightarrow 2 x^{3} - 3 x^{2} - 3 x + 2 = 2 x^{3} + b x^{2} - c x + d$
Comparing the coefficients, we get
$b = - 3, c = 3, d = 2$
$∴ b = - 3, c = 3, d = 2$

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The expression 2x3+bx2−cx+d leaves the same remainder, when divided by x+1 or x−2 or 2x−1. Find b and c.

(x+1)(x−2)(2x−1)=2x3+bx2−cx+d⇒(x2−2x+x−2)(2x−1)=2x3+bx2−cx+d⇒(x2−x−2)(2x−1)=2x3+bx2−cx+d⇒2x3−x2−2x2+x−4x+2=2x3+bx2−cx+d⇒2x3−3x2−3x+2=2x3+bx2−cx+dComparing the coefficients, we getb=−3,c=3,d=2∴b=−3,c=3,d=2

The expression $2 x^{3} + b x^{2} - c x + d$ leaves the same remainder, when divided by $x + 1$ or $x - 2$ or $2 x - 1$ . Find $b$ and $c$ .