
Verify division algorithm i.e, Dividend=Divisor×Quotient+Remainder, in each of the following:
I) \[14{x^2} + 13x - 15\] by \[7x - 4\]
II) \[15{z^2} - 20{z^2} - 13z - 12\] by \[3z - 6\]
III) \[34x - 22{x^3} - 12{x^4} - 10{x^2} - 75\] by \[3x + 7\]
IV) \[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\] by \[3y - 2\]
V) \[6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6\] by \[2{y^3} + 1\]
VI) \[4{y^3} + 8{y^2} + 8y + 7\] by \[2{y^2} - y + 1\]
Answer
587.7k+ views
Hint: The polynomial division algorithm is the method of dividing the terms that consists of polynomials with variables x, y or something else. We will follow the polynomial division algorithm to solve the equations.
Complete step-by-step answer:
i) \[14{x^2} + 13x - 15\] by \[7x - 4\]
It is given that the dividend is D=\[14{x^2} + 13x - 15\].
The divisor is given as d=\[7x - 4\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
$ \begin{array}{l}
7x - 4\mathop{\left){\vphantom{1{14{x^2} + 13x - 15}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{14{x^2} + 13x - 15}}}}
\limits^{\displaystyle \,\,\, {2x + 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\;\;\;\;\;\;\;\;\;\dfrac{{\,\,\,14{x^2}\, - 8x\;\;\;\;\;\;\;\;\;\;}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21x - 15\\
\dfrac{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21x - 12}}{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3}}
\end{array}}
\end{array} $
On dividing we get quotient as \[2x + 3\] and remainder as -3.
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {7x - 4} \right)\left( {2x + 3} \right) - 3\\
= 14{x^2} + 13x - 15
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
ii) \[15{z^3} - 20{z^2} - 13z - 12\,\]by \[3z - 6\]
The dividend is \[D = 15{z^3} - 20{z^2} - 13z - 12\,\].
The divisor is given as \[d = 3z - 6\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
3z - 6\mathop{\left){\vphantom{1{15{z^3} - 20{z^2} - 13z - 12\,}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{15{z^3} - 20{z^2} - 13z - 12\,}}}}
\limits^{\displaystyle \,\,\, {5{z^2} + \dfrac{{10}}{3}z + \dfrac{{20}}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\;\;\;\;\;\;\;\;\;\dfrac{{15{z^3}\, - 30{z^2}\;\;\;\;\;\;\;\;\;\;}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{z^2} - 13z\\
\dfrac{{\,\,\,\,\,10{z^2} - 20z}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7z - 12\\
\dfrac{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{z^2} - 20z}}{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\end{array}}
\end{array}}
\end{array}\]
On dividing we get quotient as \[5{z^2} + \dfrac{{10}}{3}z + \dfrac{{20}}{3}\] and remainder as -2.
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {3z - 6} \right)\left( {5{z^2} + \dfrac{{10}}{3}z + \dfrac{{20}}{3}} \right) - 2\\
= 15{z^3} - 20{z^2} - 13z - 12\,
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
iii) \[34x - 22{x^3} - 12{x^4} - 10{x^2} - 75\]by \[3x + 7\]
The dividend is \[34x - 22{x^3} - 12{x^4} - 10{x^2} - 75\].
The divisor is given as \[3x + 7\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
x - 3\mathop{\left){\vphantom{1{ - 12{x^4} - 22{x^3} - 10{x^2} + 34x - 75}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{ - 12{x^4} - 22{x^3} - 10{x^2} + 34x - 75}}}}
\limits^{\displaystyle \,\,\, { - 4{x^3} + 2{x^2} - 8x + 30}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 12{x^4} - 28{x^3}}}{\begin{array}{l}
\,6{x^3} - 10{x^2}\\
\dfrac{{6{x^3} + 14{x^2}}}{\begin{array}{l}
- 24{x^2} + 34x\\
\dfrac{{ - 24{x^2} - 56x}}{\begin{array}{l}
90x + 75\\
\dfrac{{90x + 210}}{{ - 135}}
\end{array}}
\end{array}}
\end{array}}
\end{array}\]
On dividing we get quotient as \[ - 4{x^3} + 2{x^2} - 8x + 30\] and remainder as -135.
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {3x + 7} \right)\left( { - 4{x^3} + 2{x^2} - 8x + 30} \right) + - 135\\
= 34x - 22{x^3} - 12{x^4} - 10{x^2} - 75
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
iv) \[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\] by \[3y - 2\]
The dividend is \[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\].
The divisor is \[3y - 2\].
iv)\[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\]by \[3y - 2\]
It is given that the dividend is D=\[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\].
The divisor is given as d=\[3y - 2\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
3y - 2\mathop{\left){\vphantom{1{15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6}}}}
\limits^{\displaystyle \,\,\, {5{y^3} - 2{y^2} + \dfrac{{13}}{3}y - \dfrac{{16}}{9}}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 15{y^4} - 10{y^3}}}{\begin{array}{l}
\,6{y^3} - 9{y^2}\\
\dfrac{{\,6{y^3} + 4{y^2}}}{\begin{array}{l}
- 13{y^2} + \dfrac{{10y}}{3} + 6\\
\dfrac{{ - 13{y^2} + \dfrac{{26y}}{3}}}{\begin{array}{l}
\dfrac{{ - 16y}}{3} + 6\\
\dfrac{{\dfrac{{16y}}{3} + \dfrac{{32}}{9}}}{{\dfrac{{22}}{9}}}
\end{array}}
\end{array}}
\end{array}}
\end{array}\]
On dividing we get quotient as \[5{y^3} - 2{y^2} + \dfrac{{13}}{3}y - \dfrac{{16}}{9}\] and remainder as \[\dfrac{{22}}{9}\].
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {3y - 2} \right)\left( {5{y^3} - 2{y^2} + \dfrac{{13}}{3}y - \dfrac{{16}}{9}} \right) + \dfrac{{22}}{9}\\
= 15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
v) \[6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6\] by \[2{y^3} + 1\]
It is given that the dividend is \[6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6\].
The divisor is given as \[2{y^3} + 1\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
3y - 2\mathop{\left){\vphantom{1{6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6}}}}
\limits^{\displaystyle \,\,\, {3{y^2} + 2y + 2}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 6{y^5} + 3{y^2}}}{\begin{array}{l}
4{y^4} + 4{y^3} + 4{y^2} + 27y + 6\\
\dfrac{{ - 4{y^4} + 0{y^3} + 0{y^2} + 2y}}{\begin{array}{l}
4{y^3} + 4{y^2} + 25y + 6\\
\dfrac{{ - 4{y^3} + 0{y^2} + 0y + 2}}{{4{y^2} + 25y + 4}}
\end{array}}
\end{array}}
\end{array}\]
On dividing we get a quotient as \[3{y^2} + 2y + 2\]and remainder as \[4{y^2} + 25y + 4\].
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {2{y^3} + 1} \right)\left( {3{y^2} + 2y + 2} \right) + 4{y^2} + 25y + 4\\
= 6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
vi) \[4{y^3} + 8{y^2} + 8y + 7\] by \[2{y^2} - y + 1\]
It is given that the dividend is \[4{y^3} + 8{y^2} + 8y + 7\].
The divisor is given as \[2{y^2} - y + 1\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
2{y^2} - y + 1\mathop{\left){\vphantom{1{4{y^3} + 8{y^2} + 8y + 7}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{4{y^3} + 8{y^2} + 8y + 7}}}}
\limits^{\displaystyle \,\,\, {2y + 5}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 4{y^3} - 8{y^2} + 2y}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{y^2} + 6y + 7\\
\dfrac{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 10{y^2} - 5y + 5}}{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11y + 2}}
\end{array}}\,\,\,\,\,\,\,\,\,\,\,\,\,
\end{array}\]
we get quotient as \[2y + 5\] and remainder as \[11y + 2\].
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {2{y^2} - y + 1} \right)\left( {2y + 5} \right) + 11y + 2\\
= 4{y^3} + 8{y^2} + 8y + 7
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
Note: Make sure that the polynomials in the process are sensitive and even a small mistake of placing the sign will lead to the wrong answer and collapse the whole procedure. So after division if you get the quotient and remainder then substitute the values to find whether the answer is correct or wrong.
Complete step-by-step answer:
i) \[14{x^2} + 13x - 15\] by \[7x - 4\]
It is given that the dividend is D=\[14{x^2} + 13x - 15\].
The divisor is given as d=\[7x - 4\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
$ \begin{array}{l}
7x - 4\mathop{\left){\vphantom{1{14{x^2} + 13x - 15}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{14{x^2} + 13x - 15}}}}
\limits^{\displaystyle \,\,\, {2x + 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\;\;\;\;\;\;\;\;\;\dfrac{{\,\,\,14{x^2}\, - 8x\;\;\;\;\;\;\;\;\;\;}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21x - 15\\
\dfrac{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21x - 12}}{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3}}
\end{array}}
\end{array} $
On dividing we get quotient as \[2x + 3\] and remainder as -3.
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {7x - 4} \right)\left( {2x + 3} \right) - 3\\
= 14{x^2} + 13x - 15
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
ii) \[15{z^3} - 20{z^2} - 13z - 12\,\]by \[3z - 6\]
The dividend is \[D = 15{z^3} - 20{z^2} - 13z - 12\,\].
The divisor is given as \[d = 3z - 6\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
3z - 6\mathop{\left){\vphantom{1{15{z^3} - 20{z^2} - 13z - 12\,}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{15{z^3} - 20{z^2} - 13z - 12\,}}}}
\limits^{\displaystyle \,\,\, {5{z^2} + \dfrac{{10}}{3}z + \dfrac{{20}}{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\;\;\;\;\;\;\;\;\;\dfrac{{15{z^3}\, - 30{z^2}\;\;\;\;\;\;\;\;\;\;}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{z^2} - 13z\\
\dfrac{{\,\,\,\,\,10{z^2} - 20z}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7z - 12\\
\dfrac{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{z^2} - 20z}}{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\end{array}}
\end{array}}
\end{array}\]
On dividing we get quotient as \[5{z^2} + \dfrac{{10}}{3}z + \dfrac{{20}}{3}\] and remainder as -2.
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {3z - 6} \right)\left( {5{z^2} + \dfrac{{10}}{3}z + \dfrac{{20}}{3}} \right) - 2\\
= 15{z^3} - 20{z^2} - 13z - 12\,
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
iii) \[34x - 22{x^3} - 12{x^4} - 10{x^2} - 75\]by \[3x + 7\]
The dividend is \[34x - 22{x^3} - 12{x^4} - 10{x^2} - 75\].
The divisor is given as \[3x + 7\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
x - 3\mathop{\left){\vphantom{1{ - 12{x^4} - 22{x^3} - 10{x^2} + 34x - 75}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{ - 12{x^4} - 22{x^3} - 10{x^2} + 34x - 75}}}}
\limits^{\displaystyle \,\,\, { - 4{x^3} + 2{x^2} - 8x + 30}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 12{x^4} - 28{x^3}}}{\begin{array}{l}
\,6{x^3} - 10{x^2}\\
\dfrac{{6{x^3} + 14{x^2}}}{\begin{array}{l}
- 24{x^2} + 34x\\
\dfrac{{ - 24{x^2} - 56x}}{\begin{array}{l}
90x + 75\\
\dfrac{{90x + 210}}{{ - 135}}
\end{array}}
\end{array}}
\end{array}}
\end{array}\]
On dividing we get quotient as \[ - 4{x^3} + 2{x^2} - 8x + 30\] and remainder as -135.
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {3x + 7} \right)\left( { - 4{x^3} + 2{x^2} - 8x + 30} \right) + - 135\\
= 34x - 22{x^3} - 12{x^4} - 10{x^2} - 75
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
iv) \[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\] by \[3y - 2\]
The dividend is \[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\].
The divisor is \[3y - 2\].
iv)\[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\]by \[3y - 2\]
It is given that the dividend is D=\[15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6\].
The divisor is given as d=\[3y - 2\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
3y - 2\mathop{\left){\vphantom{1{15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6}}}}
\limits^{\displaystyle \,\,\, {5{y^3} - 2{y^2} + \dfrac{{13}}{3}y - \dfrac{{16}}{9}}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 15{y^4} - 10{y^3}}}{\begin{array}{l}
\,6{y^3} - 9{y^2}\\
\dfrac{{\,6{y^3} + 4{y^2}}}{\begin{array}{l}
- 13{y^2} + \dfrac{{10y}}{3} + 6\\
\dfrac{{ - 13{y^2} + \dfrac{{26y}}{3}}}{\begin{array}{l}
\dfrac{{ - 16y}}{3} + 6\\
\dfrac{{\dfrac{{16y}}{3} + \dfrac{{32}}{9}}}{{\dfrac{{22}}{9}}}
\end{array}}
\end{array}}
\end{array}}
\end{array}\]
On dividing we get quotient as \[5{y^3} - 2{y^2} + \dfrac{{13}}{3}y - \dfrac{{16}}{9}\] and remainder as \[\dfrac{{22}}{9}\].
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {3y - 2} \right)\left( {5{y^3} - 2{y^2} + \dfrac{{13}}{3}y - \dfrac{{16}}{9}} \right) + \dfrac{{22}}{9}\\
= 15{y^4} - 16{y^3} - 9{y^2} - \dfrac{{10}}{3}y + 6
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
v) \[6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6\] by \[2{y^3} + 1\]
It is given that the dividend is \[6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6\].
The divisor is given as \[2{y^3} + 1\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
3y - 2\mathop{\left){\vphantom{1{6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6}}}}
\limits^{\displaystyle \,\,\, {3{y^2} + 2y + 2}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 6{y^5} + 3{y^2}}}{\begin{array}{l}
4{y^4} + 4{y^3} + 4{y^2} + 27y + 6\\
\dfrac{{ - 4{y^4} + 0{y^3} + 0{y^2} + 2y}}{\begin{array}{l}
4{y^3} + 4{y^2} + 25y + 6\\
\dfrac{{ - 4{y^3} + 0{y^2} + 0y + 2}}{{4{y^2} + 25y + 4}}
\end{array}}
\end{array}}
\end{array}\]
On dividing we get a quotient as \[3{y^2} + 2y + 2\]and remainder as \[4{y^2} + 25y + 4\].
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {2{y^3} + 1} \right)\left( {3{y^2} + 2y + 2} \right) + 4{y^2} + 25y + 4\\
= 6{y^5} + 4{y^4} + 4{y^3} + 7{y^2} + 27y + 6
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
vi) \[4{y^3} + 8{y^2} + 8y + 7\] by \[2{y^2} - y + 1\]
It is given that the dividend is \[4{y^3} + 8{y^2} + 8y + 7\].
The divisor is given as \[2{y^2} - y + 1\].
We know the equation to verify the division algorithm which is given as,
\[D = d \times q + r\]
Now, we will use the dividing method, when we divide both the terms,
\[\begin{array}{l}
2{y^2} - y + 1\mathop{\left){\vphantom{1{4{y^3} + 8{y^2} + 8y + 7}}}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{{4{y^3} + 8{y^2} + 8y + 7}}}}
\limits^{\displaystyle \,\,\, {2y + 5}}\\
\;\;\;\;\;\;\;\;\;\dfrac{{ - 4{y^3} - 8{y^2} + 2y}}{\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{y^2} + 6y + 7\\
\dfrac{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 10{y^2} - 5y + 5}}{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11y + 2}}
\end{array}}\,\,\,\,\,\,\,\,\,\,\,\,\,
\end{array}\]
we get quotient as \[2y + 5\] and remainder as \[11y + 2\].
On substituting the values in the above equation then we will get the Dividend as,
\[\begin{array}{c}
= \left( {2{y^2} - y + 1} \right)\left( {2y + 5} \right) + 11y + 2\\
= 4{y^3} + 8{y^2} + 8y + 7
\end{array}\]
Therefore, the given equation is verified using the division algorithm.
Note: Make sure that the polynomials in the process are sensitive and even a small mistake of placing the sign will lead to the wrong answer and collapse the whole procedure. So after division if you get the quotient and remainder then substitute the values to find whether the answer is correct or wrong.
Recently Updated Pages
Master Class 8 Social Science: Engaging Questions & Answers for Success

Master Class 8 English: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 Maths: Engaging Questions & Answers for Success

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Full form of STD, ISD and PCO

Advantages and disadvantages of science

Right to vote is a AFundamental Right BFundamental class 8 social science CBSE

What are the 12 elements of nature class 8 chemistry CBSE


