Question

what must be subtracted from (4x⁴-2x³-6x²+2x+6)so that the result is exactly divisible by (2x²+x-1) ?​

Answer 1

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$\huge{\underline{\underline{\sf{\blue{Solution:-}}}}}$

when the given polynomial is divided by a quadratic polynomial, then the remainder is a linear expression, say (ax+b).

$let \: p(x) =( {4x}^{4} - {2x}^{3} - {6x}^{2} + 2x + 6) - (ax + b) \: and \: g(x) = {2x}^{2} + x - 1.$

${\blue{\sf\underline{Then,}}}$

$p(x) = {4x}^{4} - {2x}^{3} - {6x}^{2} + (2 - a)x + (6 - b)$

$and \: g(x) = ( {2x}^{2} + x - 1) = ( {2x}^{2} + 2x - x - 1) = 2x(x + 1) - (x + 1) \\ = (x + 1)(2x - 1).$

Now,p(x) will be divisible by g(x) only when it is divisible by (x+1) as well as by (2x-1).

$now \: (x + 1 = 0 = x = - 1) \: and \: (2x - 1 = 0 = 2x = 1 = x = \frac{1}{2} ).$

By factor theorem, p(x) will be divisible by g (x),if p (-1)=0 and $p( \frac{1}{2} ) = 0$

$p( - 1) = 0</em><em> </em><em>⇒</em><em> 4 \times {( - 1)}^{4} - 2 \times { ( - 1)}^{3} - 6 \times {( - 1)}^{2} + (2 - a) \times ( - 1) + 6 - b = 0$

$</em><em>⇒</em><em> 4 + 2 - 6 - 2 + a + 6 - b = 0 = a - b = - 4 \: \: \: \: \: \: \:..... (i)$

$p( \frac{1}{2} ) = 0 </em><em>⇒</em><em> 4 {( \frac{1}{2}) }^{4} - 2 \times {( \frac{1}{2} )}^{3} - 6 \times {( \frac{1}{2} )}^{2} + 2 - a) \times \frac{1}{2} + (6 - b) = 0 \\ </em><em>⇒</em><em> (4 \times \frac{1}{16} ) - (2 \times \frac{1}{8} ) - (6 \times \frac{1}{4} ) + 1 - \frac{a}{2} + 6 - b = 0 \\ </em><em>⇒</em><em> \frac{1}{4} - \frac{1}{4} - \frac{3}{2} + 1 - \frac{a}{2} + 6 - b = 0 </em><em>⇒</em><em> \frac{a}{2} + b = \frac{11}{2} \\ </em><em>⇒</em><em>a + 2b = 11. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ........(ii)$

on solving (i) and (ii), we get a = 1 and b = 5.

hence, the required expression to be subtracted is (x+5)

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