Question

$(4l^5-6l^4+8l^3)÷2l^2$

• A. $(2l^2-3l+4)$
• B. $l(2l^2-3l+4)$
• C. $l(2l^2-3l+3)$
• D. None

Answer 1

The correct option is B $l(2l^2-3l+4)$

We divide the given polynomial $4l^5-6l^4+8l^3$ by the monomial $2l^2$ as shown below:

$\frac{4l^5-6l^4+8l^3}{2l^2}=\frac{4l^5}{2l^2}-\frac{6l^4}{2l^2}+\frac{8l^3}{2l^2}=\left(\frac{2\times 2\times l^5}{2l^2}\right)-\left(\frac{2\times 3\times l^4}{2l^2}\right)+\left(\frac{2\times 2\times 2\times l^3}{2l^2}\right)=\left(\frac{2\times l^5\times l^{-2}}{1}\right)-\left(\frac{3\times l^4\times l^{-2}}{1}\right)+\left(\frac{2\times 2\times l^3\times l^{-2}}{1}\right)$

$=(2\times l^{(5-2)})-(3\times l^{(4-2)})+(4\times l^{(3-2)})(\because a^x+a^y=a^{x+y})=2l^3-3l^2+4l=l(2l^2-3l+4)$

Hence, $\frac{4l^5-6l^4+8l^3}{2l^2}=l(2l^2-3l+4)$.