# Divide:(i) $acx^2+(bc+ad)x+bd$ by $ax+b$(ii) $(a^2+2ab+b^2)-(a^2+2ac+c^2)$ by $2a+b+c$

Given:

The given expressions are:

(i) $acx^2+(bc+ad)x+bd$ by $ax+b$

(ii) $(a^2+2ab+b^2)-(a^2+2ac+c^2)$ by $2a+b+c$

To do:

We have to divide the given expressions.

Solution:

We have to divide the given polynomials by simplifying them using algebraic formulas.

Polynomials:

Polynomials are expressions in which each term is a constant multiplied by a variable raised to a whole number power.

Therefore,

(i) The given expression is $acx^2+(bc+ad)x+bd$ by $ax+b$.

$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{acx^2+(bc+ad)x+bd}{ax+b}$

$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{acx^2+bcx+adx+bd}{ax+b}$

$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{cx(ax+b)+d(ax+b)}{ax+b}$        (Taking $cx$ and $d$ common)

$acx^2+(bc+ad)x+bd \div (ax+b)=\frac{cx(ax+b)}{ax+b}+\frac{d(ax+b)}{ax+b}$

$acx^2+(bc+ad)x+bd \div (ax+b)=cx+d$

Hence, $acx^2+(bc+ad)x+bd$ divided by $ax+b$ is $cx+d$.

(ii) The given expression is $(a^2+2ab+b^2)-(a^2+2ac+c^2)$ by $2a+b+c$.

$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=\frac{(a^2+2ab+b^2)-(a^2+2ac+c^2)}{2a+b+c}$

$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=\frac{(a+b)^2-(a+c)^2}{2a+b+c}$                [Since $(x+y)^2=x^2+2xy+y^2$]

$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=\frac{(a+b+a+c)(a+b-a-c)}{2a+b+c}$                 [Since $x^2-y^2=(x+y)(x-y)$]

$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=\frac{(2a+b+c)(b-c)}{2a+b+c}$

$(a^2+2ab+b^2)-(a^2+2ac+c^2) \div (2a+b+c)=b-c$

Hence, $(a^2+2ab+b^2)-(a^2+2ac+c^2)$ divided by $2a+b+c$ is $b-c$.

Updated on: 13-Apr-2023

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