Question
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder :
$(y^3 – 3y^2 + 5y – 1) ÷ (y – 1)$
$(y^3 – 3y^2 + 5y – 1) ÷ (y – 1)$
✓
Answer
Synthetic division:
$\left(y^3-3 y^2+5 y-1\right) \div(y-1)$
Dividend $=y^3-3 y^2+5 y-1$
Coefficient form of the dividend $=(1,-3,5,-1)$
Divisor $= y -1$
$\therefore$ Opposite of $-1$ is $1$ .
$\therefore$ Coefficient form of quotient $=(1,-2,3)$
$\therefore$ Quotient $= y ^2-2 y +3$,
Remainder $=2$
Linear division method:
$y^3-3 y^2+5 y-1$
To get the term $y ^3$, multiply $( y -1)$ by $y ^2$ and add $y ^2$
$=y^2(y-1)+y^2-3 y^2+5 y-1$
$=y^2(y-1)-2 y^2+5 y-1$
To get the term $-2 y ^2$, multiply $( y -1$ ) by -2 y and subtract 2 y ,
$=y^2(y-1)-2 y(y-1)-2 y+5 y-1$
$=y^2(y-1)-2 y(y-1)+3 y-1$
To get the term $3 y$, multiply $(y-1)$ by 3 and add 3 ,
$=y^2(y-1)-2 y(y-1)+3(y-1)+3-1$
$=(y-1)\left(y^2-2 y+3\right)+2$
$\therefore \text { Quotient }=y^2-2 y+3$
$\text { Remainder }=2$
$\left(y^3-3 y^2+5 y-1\right) \div(y-1)$
Dividend $=y^3-3 y^2+5 y-1$
Coefficient form of the dividend $=(1,-3,5,-1)$
Divisor $= y -1$
$\therefore$ Opposite of $-1$ is $1$ .
$\therefore$ Coefficient form of quotient $=(1,-2,3)$
$\therefore$ Quotient $= y ^2-2 y +3$,
Remainder $=2$
Linear division method:
$y^3-3 y^2+5 y-1$
To get the term $y ^3$, multiply $( y -1)$ by $y ^2$ and add $y ^2$
$=y^2(y-1)+y^2-3 y^2+5 y-1$
$=y^2(y-1)-2 y^2+5 y-1$
To get the term $-2 y ^2$, multiply $( y -1$ ) by -2 y and subtract 2 y ,
$=y^2(y-1)-2 y(y-1)-2 y+5 y-1$
$=y^2(y-1)-2 y(y-1)+3 y-1$
To get the term $3 y$, multiply $(y-1)$ by 3 and add 3 ,
$=y^2(y-1)-2 y(y-1)+3(y-1)+3-1$
$=(y-1)\left(y^2-2 y+3\right)+2$
$\therefore \text { Quotient }=y^2-2 y+3$
$\text { Remainder }=2$
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