Question

What must be added to (3x^3+x^2-22x+9) so that the result is completely divisible by (3x^2+7x-6)
Answer-(2x+3).Please show full process

Answer 1

Hello Mate!

Please refer to attatchment

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Answer 2

Answer:

$3x^2+9x-3$

Step-by-step explanation:

Dividend = $(3x^3+x^2-22x+9)$

Divisor = $(3x^2+7x-6)$

$Dividend = (Divisor \times Quotient)+Remainder$

$(3x^3+x^2-22x+9) = ((3x^2+7x-6) \times x-2)+(-2x-3)$

remainder = (-2x-3)

Divisor - Remainder =$(3x^2+7x-6) - (-2x-3)=3x^2+7x-6+2x+3=3x^2+9x-3$

Hence $3x^2+9x-3$ must be added to $(3x^3+x^2-22x+9)$so that the result is completely divisible by  $(3x^2+7x-6)$