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Last updated at March 16, 2023 by Teachoo
Example 15 Divide 44 (π₯^4 β 5π₯^3 β 24π₯^2) by 11x (x β 8) We first factorize 44 (π₯^4 β 5π₯^3 β 24π₯^2) = 44 (π₯^2 Γ π₯^2 β 5π₯ Γ π₯^2 β 24 Γ π₯^2) Taking π₯^2 common, = γ44π₯γ^2 (π₯^2 β 5π₯ β 24) By middle term splitting, = 44π₯^2 (π₯^2+3π₯ β 8π₯ β 24) = 44π₯^2 [(π₯^2+3π₯) β (8π₯+24)] Both have x as common factor Both have 8 as common factor Splitting the middle term We need to find two numbers whose Sum = β5 Product = β24 So, we write β5x = 3x β 8x = 44π₯^2 [π₯(π₯+3) β8(π₯+3)] Taking π₯+3 common, = 44π₯^2 (π₯+3) (π₯ β 8) Dividing (44 (π₯^4 β 5π₯^3 β 24π₯^2 ))/(11π₯ (π₯ β 8)) = (44 π₯^2 (π₯ + 3) (π₯ β 8))/(11π₯ (π₯ β 8)) = 44/11 Γ π₯^2/π₯ Γ (π₯ + 3) Γ ((π₯ β 8))/((π₯ β 8)) = 4 Γ π₯ Γ (π₯ + 3) = 4π (π + 3)