Find the quotient and the remainder
and also verify the result
x3 + 2x - 52-3 is divided by 32+3x2
Answers
Step-by-step explanation:
\textbf{Given:}Given:
30x^4+11x^3-82x^2+12x+48{\div}3x^2+2x-430x
4
+11x
3
−82x
2
+12x+48÷3x
2
+2x−4
\textbf{To find:}To find:
\text{Quotient and remainder}Quotient and remainder
\textbf{Solution:}Solution:
\text{Consider,}Consider,
\begin{gathered}\begin{array}{r|l}&10x^2-3x-12\\\cline{2-2}3x^2+2x-4&30x^4+11x^3-82x^2+12x+48\\&30x^4+20x^3-40x^2\\\cline{2-2}&****-9x^3-42x^2+12x\\&****-9x^3-\;6x^2+12x\\\cline{2-2}&********-36x^2+\;0x+48\\&********-36x^2-12x+48\\\cline{2-2}\\&**************12x\\\cline{2-2}\end{array}\end{gathered}
\cline2−23x
2
+2x−4
\cline2−2
\cline2−2
\cline2−2
\cline2−2
10x
2
−3x−12
30x
4
+11x
3
−82x
2
+12x+48
30x
4
+20x
3
−40x
2
∗∗∗∗−9x
3
−42x
2
+12x
∗∗∗∗−9x
3
−6x
2
+12x
∗∗∗∗∗∗∗∗−36x
2
+0x+48
∗∗∗∗∗∗∗∗−36x
2
−12x+48
∗∗∗∗∗∗∗∗∗∗∗∗∗∗12x
\text{From the above long division,}From the above long division,
\text{Quotient}=10x^2-3x-12Quotient=10x
2
−3x−12
\text{Remainder}=12xRemainder=12x
\textbf{Verification:}Verification:
\boxed{\textbf{Dividend=(Divisor$\times$Quotient)+Remainder}}
Dividend=(Divisor×Quotient)+Remainder
\textbf{(Divisor$\times$Quotient)+Remainder}(Divisor×Quotient)+Remainder
=(3x^2+2x-4)(10x^2-3x-12)+12x=(3x
2
+2x−4)(10x
2
−3x−12)+12x
=(30x^4-9x^3-36x^2+20x^3-6x^2-24x-40x^2+12x+48)+12x=(30x
4
−9x
3
−36x
2
+20x
3
−6x
2
−24x−40x
2
+12x+48)+12x
=(30x^4+11x^3-82x^2+0x+48)+12x=(30x
4
+11x
3
−82x
2
+0x+48)+12x
=30x^4+11x^3-82x^2+12x+48=30x
4
+11x
3
−82x
2
+12x+48
=\textbf{Dividend}=Dividend
\textbf{Hence verified}Hence verified