Question

ax^3+3x^2-13and 2x^3-5x+a when divided by x-2 leave same remainder. Find value of a

Answer 1

AnswEr :

$\bf{\Large{\underline{\sf{Given\::}}}}$

The polynomials ax³ + 3x² -13 and 2x³ - 5x + a when divided by x-2 leave same remainder.

$\bf{\Large{\underline{\sf{To\:find\::}}}}$

The value of a.

$\bf{\Large{\underline{\rm{\orange{Explanation\::}}}}}$

$\bf{\red{We\:have}\begin{cases}\sf{f(x)=ax^{3} +3x^{2} -13}\\ \sf{g(x)=2x^{3} -5x+a}\end{cases}}$

A/q

$\leadsto\sf{x-2=0}\\\\\leadsto\sf{\pink{x=2}}$

So,

$|\Rightarrow\tt{f(2)=g(2)}\\\\\\|\Rightarrow\tt{a.(2)^{3} +3(2)^{2} -13=2(2)^{3} -5(2)+a}\\\\\\|\Rightarrow\tt{a.8+3*4-13=2.8-5*2+a}\\\\\\|\Rightarrow\tt{8a+12-13=16-10+a}\\\\\\|\Rightarrow\tt{8a-1=6+a}\\\\\\|\Rightarrow\tt{8a-a=6+1}\\\\\\|\Rightarrow\tt{7a=7}\\\\\\|\Rightarrow\tt{a\:=\:\cancel{\dfrac{7}{7} }}\\\\\\|\Rightarrow\tt{\red{a=1}}$

Thus,

$\bigstar$The value of a is 1.

Answer 2

$\huge \boxed{ \fcolorbox{cyan}{grey}{answer : }}$

let

$\rm{p1(x) = a {x}^{3} + 3 {x}^{2} - 13}$

and

$\rm{p2(x) = 2 {x}^{3} - 5x + a}$

$\sf{according \: to \: the \: equation}$

$\rm{p1( 2) \: and \: p2( 2)}$

$\sf{p1( 2) = a {( 2)}^{2} + 3 {( 2)}^{2} - 13}$

$\sf{ 8a + 12 - 13}$

$\sf{2 {( 2)}^{2} + 5( 2) + a}$

$\sf{ 16 + 10 + a}$

then

8a - a = 16 -10 -12 + 13

$\rm{ 7a - 1 = - 6}$

7a = 6+1

7a = 7

a = 1

$\bf{ \huge{ \boxed{ \red{ \tt{a = 1 \: }}}}}}$