Question

Divide 6x^3+13x^2+x-2 by 2x + 1,find quotient and remainder​

Answer 1

Solution :

Division of Polynomial!

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$\sf{p(x) = 6 {x}^{3} + 13 {x}^{2} + x - 2}$

$\sf{g(x) = 2x + 1}$

From Long Division Method,

Quotient obtained is :

$\sf{Quotient = 3 {x}^{2} + 5x - 2}$

$\sf{Remainder= 0}$

Refer to the attachment!

Verification :

$\sf{Dividend = Divisor \times Quotient + Remainder}$

$\longrightarrow\:\sf{6 {x}^{3} + 13 {x}^{2} + x - 2 = (2x + 1)(3 {x}^{2} + 5x - 2) + 0}$

$\longrightarrow\:\sf{6 {x}^{3} + 13 {x}^{2} + x - 2 =2x(3 {x}^{2} + 5x - 6) + 1(3 {x}^{2} + 5x - 6)}$

$\longrightarrow\:\sf{6 {x}^{3} + 13 {x}^{2} + x - 2 = 6 {x}^{3} +10 {x}^{2} - 4x + 3 {x}^{2} + 5x - 2}$

$\longrightarrow\:\sf{6 {x}^{3} + 13 {x}^{2} + x - 2 = 6 {x}^{3} +13 {x}^{2} + x - 2}$

Hence, Verified!

Answer 2

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❚ QuEstiOn ❚

# Divide (6x^3+13x^2+x-2) by (2x + 1) find quotient and remainder .

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❚ ANsWeR ❚

✺ Given :

• dividend = 6x^3+13x^2+x-2
• divisor = 2x + 1

✺ To FinD :

• quotient = ?
• remainder = ?

✺ Explanation :

✏ By L.DM. method :-

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So from here ,

❏ we can see that the remainder is = O

❏ we can see that the quotient is = (3x²+5x-2)

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✏ By Remainder Theorm :-

# f(x) = 6x^3+13x^2+x-2

# g(x) = 2x+1

Now , applying Remainder theorm ,

➩ 2x+1= 0

➩ 2x = -1

➩ x = $\dfrac{-1}{2}$

∴ Remainder = f( $\dfrac{-1}{2}$ )

$\implies Remainder = 6(\dfrac{-1}{2})^3+13(\dfrac{-1}{2})^2+(\dfrac{-1}{2})-2$

$\implies Remainder =\dfrac{-6}{8}+\dfrac{13}{4}-\dfrac{1}{2})-2$

$\implies Remainder = \dfrac{-6+26-4-16}{8}$

➩ Remainder = 0

Now ,

∴ dividend = ( divisor × quotient ) + remainder

➩ 6x^3+13x^2+x-2 = (2x+1)×quotient + 0

➩ 6x^3+13x^2+x-2 = (2x+1)×quotient

$\implies quotient =\dfrac{ 6x^3+13x^2+x-2}{2x+1}$

$\implies quotient =\dfrac{3x^2(2x+1)+5x(2x+1)-2(2x+1)}{(2x+1)}$

$\implies quotient =\dfrac{\cancel{(2x+1)}(3x^2+5x-2)}{\cancel{(2x+1)}}$

➩quotient = 3x²+5x-2

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