Question

the expression 2x³ + bx² - cx + d leave the same remainder when divided by X + 1 or x - 2 or 2x- 1 find b and c
it is from ch -3 class 10th
u can use elimination method, substitution method or cross multiplication method

the best answer will be marked brainest

Answer 1

$\huge{\mathfrak{Solution:-}}$

$\sf{Let\;p(x) = 2x^{3}+bx^{2}-cx+d}$

$\sf{When\;p(x)\;is\;divided\;by\;(x+1),\;then\;remainder\;will\;be\;p(-1)}$

$\sf{Here,\;p(-1)=2(-1)^{3}+b(-1)^{2}-c(-1)+d}\\ \\ \sf{= -2+b+c+d}$

$\sf{When\;p(x)\;is\;divided\;by\;x-2,\;then\;remainder\;will\;be\;p(2)}$

$\sf{Here,\;p(2) = 2(2)^{3}+b(2)^{2}-c(2)+d}\\ \\ \sf{= 16+4b-2c+d}$

$\sf{When\;p(x)\;is\;divided\;by\;2x-1,\;then\;remainder\;will\;be\;p(\frac{1}{2})}$

$\sf{Here,\;p(\frac{1}{2})=2(\frac{1}{2})^{3}+b(\frac{1}{2})^{2}-c(\frac{1}{2})+d}\\ \\ \sf{=\frac{1}{4}+\frac{b}{4}-\frac{c}{2}+d}$

$\sf{Now,\;remainder\;in\;all\;three\;cases\;are\;same}$

$\sf{\therefore -2+b+c+d=16+4b-2c+d}$

$\sf{\implies -2+b+c+d-16-4b+2c-d=0}$

$\sf{\implies -3b+3c-18=0}$

$\sf{\implies b-c=-6\;\;\;\;\;\;\;\;\;..........(1)}$

$\sf{Also,}$

$\sf{-2+b+c+d = \frac{1}{4}+\frac{b}{4}-\frac{c}{2}+d}$

$\sf{\implies -2+b+c+d-\frac{1}{4}-\frac{b}{4}+\frac{c}{2}-d=0}$

$\sf{\implies \frac{3b}{4}+\frac{3c}{2}-\frac{9}{4}=0}$

$\sf{\implies 3b+6c-9=0}$

$\sf{\implies b+2c-3=0}$

$\sf{\implies b+2c=3\;\;\;\;\;\;\;......(2)}$

$\sf{From\;(2)-(1),\;we\;get}$

$\sf{3c=9}$

$\sf{\implies c=3}$

$\sf{Putting\;c=3\;in\;(1),}$

$\sf{b-3=-6}$

$\sf{\implies b = -3}$

$\sf{Hence,}$

$\boxed{\sf{b=-3}}$

$\boxed{\sf{c=3}}$

$\huge{\mathcal{\red{@AASHIQ}}}$

Answer 2

$\huge\bold\\ANSWER:-}}}}}$

{hence it is x-2=0, x=2

f(2)= 2(2)^3+b(2)^2-c(2)+d

= 16+4b-2c+d }

{hence it is 2x-1 =0, x= 0.5

f(0.5)= 2(0.5)^3+b(0.5)^2-c(0.5)+d

= 0.25 + 0.25b - 0.5c+d }

{thus , remainder, conclude that:

{-2+b+c+d=16+4b-2c+d=0.25+ 0.25b - 0.5c+d }

[-2+b+c=16+4b-2c=0.25+ 0.25b - 0.5c }

[[[[2 equation]]]]]

{-2+b+c=16+4b-2c}

also.. below .

{16+4b-2c=0.25+ 0.25b - 0.5c }

From first..

{-2+b+c=16+4b-2c }

{b+c= 18+4b-2c}

{0=18+3b-3c }

{3c=18+3b }

{c= 6+b}

[[[[From second]]]

{{16+4b-2c=0.25+ 0.25b - 0.5c }}

{{64+16b-8c= 1+b-2c}

{63+15b= 6c}

{Substitute now here 3 into 4}}

{63+15b= 6( 6+b) }

{63+ 15b= 36+6b }

{9b= -27 }

{b= -3 }

[Substitute b= -3 into ]

{c= 6-3 }

{c=3 }

{Therefore,now here}


====> 2x^3+bx^2-cx+d ====> 2x^3 -3x^2-3x+ d

{Since it is x+2=0, x= -2}

{(-2) = 2x^3 -3x^2-3x+ d }

===> 2(-2)^3 -3(-2)^2-3(-2)+ d

===> -16 -12 +6+d

{As now is (x) is divisible by x+2, it implies that }

[-16 -12 +6+d = 0]

{{d=22}}

===>hence the answer is { b= -3, c=3, d= 22}

hence proved:)

hope it helps:--

T!—!ANKS!!!