Question

solve this
question for me

Answer 1

what is the sine middle of 11x^ and 82x^

Answer 2

$Let \: p(x) = 30x^{4}+11x^{3}-82x^{2}-12x-48 \:and\: g(x) = 3x^{2}+2x-4$

Quotient: 10x²-3x-12_____

3x²+2x-4|30x⁴+11x³-82x²-12x-48

************* 30x⁴+20x³-40x²

________________________

************* -9x³-42x²-12x

************* -9x³-6x² +12x

________________________

****************** -36x²-24x -48

****************** -36x²-24x+48

________________________

Remainder = - 96

________________________

Verification:

Dividend = 30x⁴+11x³-82x²-12x-48

Divisor = 3x²+2x-4,

Quotient = 10x²-3x-12

Remainder = - 96

By Division Algorithm :

$\pink { Divisor \times Quotient + Remainder }\\\pink { = Dividend }$

$Divisor \times Quotient + Remainder\\=(3x^{2}+2x-4)(10x^{2}-3x-12)+ (-96) \\= </p><p>= 3x^{2}(10x^{2}-3x-12) \\+2x(10x^{2}-3x-12)\\-4(10x^{2}-3x-12)-96 \\= 30x^{4}-9x^{3}-36x^{2} \\+20x^{3}-6x^{2}-24x\\-40x^{2}+12x+48\\+24x-96 \\= 30x^{4}+(-9+20)x^{3}\\+(-36-6-40)x^{2}+(-24+12)x\\+48-96 \\= 30x^{4} +11x^{3} -82x^{2} -12x-48\\= Dividend$

•••♪