Question

$\huge \underline { \fbox \purple{Question \: : - }}$


The expression 2x³ + ax² + bx – 2 leaves remainder 7 and 0 when divided by 2x – 3 and x + 2 respectively. Calculate the values of a and b.

Note :- Spamming = 20 relevant answers reported.

Wrong answer = 5 relevent answers reported.

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Answer 1

Answer:

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Answer 2

❍ Let's Consider p(x) = 2x³ + ax² + bx - 2

Given that ,

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

⠀$\underline {\sf{\star\:Now \: By \: Substituting \: x=\dfrac{3}{2} \: in \: p(x) \::}}$

$\therefore \sf{ 2 \bigg( \dfrac{3}{2}\bigg)^{3} + a\bigg(\dfrac{3}{2}\bigg)^{2} + b\bigg( \dfrac{3}{2}\bigg) -2 }$

• On dividing p(x) by 2x - 3 , it's leave a remainder 7 .

$:\implies \sf{ 2 \bigg( \dfrac{3}{2}\bigg)^{3} + a\bigg(\dfrac{3}{2}\bigg)^{2} + b\bigg( \dfrac{3}{2}\bigg) -2 = 7 }$

$:\implies \sf{ 2 \bigg( \dfrac{3}{2}\bigg)^{3} + a\bigg(\dfrac{3}{2}\bigg)^{2} + b\bigg( \dfrac{3}{2}\bigg) = 7+ 2 }$

$:\implies \sf{ 2 \bigg( \dfrac{3}{2}\bigg)^{3} + a\bigg(\dfrac{3}{2}\bigg)^{2} + b\bigg( \dfrac{3}{2}\bigg) = 9 }$

$:\implies \sf{ 2 \times \dfrac{27}{8} + a\times \dfrac{9}{4} + b\times \dfrac{3}{2} = 9 }$

$:\implies \sf{ \cancel {2} \times \dfrac{27}{\cancel {8}} + a\times \dfrac{9}{4} + b\times \dfrac{3}{2} = 9 }$

$:\implies \sf{ \dfrac{27}{4} + a\times \dfrac{9}{4} + b\times \dfrac{3}{2} = 9 }$

$:\implies \sf{ \dfrac{27}{4} + \dfrac{9a}{4} + \dfrac{3b}{2} = 9 }$

$:\implies \sf{ \dfrac{27 + 9a + 6b }{4} = 9 }$

$:\implies \sf{ 27 + 9a + 6b = 9\times 4 }$

$:\implies \sf{ 27 + 9a + 6b = 36 }$

$:\implies \sf{ 9a + 6b = 36 -27 }$

$:\implies \sf{ 9a + 6b = 9 }$

$:\implies \sf{ 9a + 6b -9 = 0 }$

$:\implies \sf{ \bigg(3a + 2b - 3 = 9 \bigg) }\:\: \longrightarrow Eq.1$

Given that ,

$:\implies \sf{x + 2= 0} \\\\:\implies \sf{x = -2}$

⠀$\underline {\sf{\star\:Now \: By \: Substituting \: x= -2 \: in \: p(x) \::}}$

$\therefore \sf{ 2 \bigg( -2 \bigg)^{3} + a\bigg(-2\bigg)^{2} + b\bigg( -2\bigg) -2 }$

• On dividing p(x) by x+2 , it's leave a remainder 0 .

$:\implies \sf{ 2 \bigg( -2 \bigg)^{3} + a\bigg(-2\bigg)^{2} + b\bigg( -2\bigg) -2 =0}$

$:\implies \sf{ 2 \bigg( -8 \bigg) + a\bigg(4\bigg) + b\bigg( -2\bigg) -2 =0}$

$:\implies \sf{ -16 + 4a -2b-2 =0 }$

$:\implies \sf{ \bigg(4a -2b-18 =0\bigg)\longrightarrow Eq.2}$

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⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Adding \: Eq.1 \:and\:2 \: we \: get, \::}}$

$\boxed{\begin{array}{cccc}\\ \sf{3a} & \sf{\cancel{+2b}} &\sf{-3}&\sf{=0} \\\\ \sf{4a}&\sf{\cancel{-2b}} & \sf{-18} & \sf{=0} \\\\ ---&---&--- \\\\ \sf{7a}& \:\:&\sf{-21}&\sf{=0}\end{array}}$

$:\implies \sf{7a - 21 = 0}$

$:\implies \sf{a = 3}$

⠀⠀⠀⠀⠀⠀$\underline {\sf{\star\:Now \: By \: Substituting \: the \: \: Value\:of\:a\:in\:Eq.1\: \::}}$

$:\implies \sf{3(3) +2b -3 = 0}$

$:\implies \sf{2b = -6}$

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

Hence ,

⠀⠀⠀⠀⠀$\therefore \underline { \mathrm { The\:Value \:of\:a\:is\:\bf{3} \: }}\:\bf{\bigstar}$

⠀⠀⠀⠀⠀$\therefore \underline {\mathrm { The\:Value \:of\:b\:is\:\bf{-3} \: }}\:\bf{\bigstar}$

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