Question

Dividing 6x3-11x2+7x+5 by 2x-3 we get as remainder *

Answer 1

Given :

• $\sf {p(x) = 6x^3-11x^2+7x+5}$
• $\sf {g(x)=2x-3}$

To Find :

• Remainder.

Solution :-

By Remainder theorem :

Let g (x) = 0.

$:\implies{\sf{2x-3=0}}\\ \\ :\implies{\sf{2x=0+3}}\\ \\ :\implies {\sf{2x=3}}\\ \\ :\implies {\tt{x=\dfrac{3}{2}}}$.

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$: \implies {\sf {p (x)=6x^3-11x^2+7x+5}}\\ \\ : \implies{\sf {p (\dfrac {3}{2})= 6\times{(\dfrac {3}{2})}^{3}-11\times {(\dfrac {3}{2})}^2+7\times \dfrac {3}{2} +5}}\\ \\ : \implies {\sf {p (\dfrac {3}{2})= 6\times(\dfrac {27}{8})-11\times (\dfrac {9}{4})+7\times \dfrac {3}{2} +5 }}\\ \\ : \implies {\sf {p (\dfrac {3}{2})=\dfrac{81}{4}-\dfrac {99}{4}+\dfrac {21}{2}+5}}\\ \\ :\implies{\sf{p (\dfrac {3}{2})=\dfrac {81-99+42+20}{4}}}\\ \\ : \implies{\sf{p (\dfrac {3}{2})=\dfrac {143-99}{4}}} \\ \\ : \implies {\sf {p (\dfrac {3}{2})=\dfrac{44}{4}}}\\ \\ :\implies{\boxed{\tt{p (\dfrac {3}{2})=11}}}$

${\therefore{\underline{\bf Remainder =11}}}$

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Answer 2

Answer:

GIVEN:

•p(x)=6x^3-11x^2+7x+5

•g(x)=2x-3

TO FIND:

The remainder (r).

THEOREM USED:

Remainder theorem .