Question
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder : $(2x^4 + 3x^3 + 4x – 2x^2) ÷ (x + 3)$
✓
Answer
Synthetic division:
$\left(2 x^4+3 x^3+4 x-2 x^2\right) \div(x+3)$
Dividend $=2 x^4+3 x^3+4 x-2 x^2$
$\therefore$ Index form $=2 x^4+3 x^3-2 x^2+4 x+0$
$\therefore$ Coefficient form of the dividend $=(2,3,-2,4,0)$
Divisor $=x+3$
$\therefore$ Opposite of $+ 3$ is $-3$
Coefficient form of quotient $=(2,-3,7,-17)$
$\therefore$ Quotient $=2 x^3-3 x^2+7 x-17$,
Remainder $=51$
Linear division method:
$2 x^4+3 x^3+4 x-2 x^2=2 x^2+3 x^3-2 x^2+4 x$
To get the term $2 x^4$, multiply $(x+3)$ by $2 x^3$ and subtract $6 x^3$,
$=2 x^3\left(x+31-6 x^3+3 x^3-2 x^2+4 x\right.$
$=2 x^3(x+3)-3 x^3-2 x^2+4 x$
To get the term $-3 x^3$, multiply $(x+3)$ by $-3 x^2$ and add $9 x^2$,
$=2 x^3(x+3)-3 x^2(x+3)+9 x^2-2 x^2+4 x$
$=2 x^3(x+3)-3 x^2(x+3)+7 x^2+4 x$
To get the term $7 x ^2$, multiply $(x+3)$ by 7 x and subtract 21 x ,
$=2 x^3(x+3)-3 x^2(x+3)+7 x(x+3)-21 x+4 x$
$=2 x^3(x+3)-3 x^2(x+3)+7 x(x+3)-17 x$
To get the term $-17 x$, multiply $(x+3)$ by -17 and add 51 ,
$=2 x^3(x+3)-3 x^2(x+3)+7 x(x+3)-17(x+3)+51$
$=(x+3)\left(2 x^3-3 x^2+7 x-17\right)+51$
$\therefore \text { Quotient }=2 x^3-3 x^2+7 x-17$
$\text { Remainder }=51$
$\left(2 x^4+3 x^3+4 x-2 x^2\right) \div(x+3)$
Dividend $=2 x^4+3 x^3+4 x-2 x^2$
$\therefore$ Index form $=2 x^4+3 x^3-2 x^2+4 x+0$
$\therefore$ Coefficient form of the dividend $=(2,3,-2,4,0)$
Divisor $=x+3$
$\therefore$ Opposite of $+ 3$ is $-3$
Coefficient form of quotient $=(2,-3,7,-17)$
$\therefore$ Quotient $=2 x^3-3 x^2+7 x-17$,
Remainder $=51$
Linear division method:
$2 x^4+3 x^3+4 x-2 x^2=2 x^2+3 x^3-2 x^2+4 x$
To get the term $2 x^4$, multiply $(x+3)$ by $2 x^3$ and subtract $6 x^3$,
$=2 x^3\left(x+31-6 x^3+3 x^3-2 x^2+4 x\right.$
$=2 x^3(x+3)-3 x^3-2 x^2+4 x$
To get the term $-3 x^3$, multiply $(x+3)$ by $-3 x^2$ and add $9 x^2$,
$=2 x^3(x+3)-3 x^2(x+3)+9 x^2-2 x^2+4 x$
$=2 x^3(x+3)-3 x^2(x+3)+7 x^2+4 x$
To get the term $7 x ^2$, multiply $(x+3)$ by 7 x and subtract 21 x ,
$=2 x^3(x+3)-3 x^2(x+3)+7 x(x+3)-21 x+4 x$
$=2 x^3(x+3)-3 x^2(x+3)+7 x(x+3)-17 x$
To get the term $-17 x$, multiply $(x+3)$ by -17 and add 51 ,
$=2 x^3(x+3)-3 x^2(x+3)+7 x(x+3)-17(x+3)+51$
$=(x+3)\left(2 x^3-3 x^2+7 x-17\right)+51$
$\therefore \text { Quotient }=2 x^3-3 x^2+7 x-17$
$\text { Remainder }=51$
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