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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 $b-c$.
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