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
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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.
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