Question

Divide the given polynomial by the given monomial.

$\left(i\right)(5x^2-6x)÷3x$

$\left(ii\right)(3y^8-4y^6+5y^4)÷y^4$

$\left(iii\right)8(x^3{y}^2{z}^2+x^2{y}^3{z}^2+x^2{y}^2{z}^3)÷4x^2{y}^2{z}^2$

$\left(iv\right)(x^3+2x^2+3x)÷2x$. [4 MARKS]

Answer 1

Solution: 1 Mark each

$\left(i\right)5x^2-6x=x(5x-6)$

$(5x^2-6x)÷3x=\frac{x(5x-6)}{3x}=\frac{1}{3}(5x-6)$


$\left(ii\right)3y^8-4y^6+5y^4=y^4(3y^4-4y^2+5)$

$(3y^8-4y^6+5y^4)÷y^4=\frac{y^4(3y^4-4y^2+5)}{y^4}=3y^4-4y^2+5$


$\left(iii\right)8(x^3{y}^2{z}^2+x^2{y}^3{z}^2+x^2{y}^2{z}^3)=8x^2{y}^2{z}^2(x+y+z)$

$8(x^3{y}^2{z}^2+x^2{y}^3{z}^2+x^2{y}^2{z}^3)÷4x^2{y}^2{z}^2=\frac{8x^2{y}^2{z}^2(x+y+z)}{4x^2{y}^2{z}^2}=2(x+y+z)$


$\left(iv\right)x^3+2x^2+3x=x(x^2+2x+3)$

$(x^3+2x^2+3x)÷2x=\frac{x(x^2+2x+3)}{2x}=\frac{1}{2}(x^2+2x+3)$