Question

factorizing difference of two squares?​

Answer 1

$\bf{Hello\:\:Mate}$

$\bf{Question\:\:1}$

$\bf{Factorize\:\:\dfrac{1}{4}b^2-49}$

$\bf{Answer}$

$\bf{We\:\:know\:\:\dfrac{1}{4}b^2\:\:=\:\:\dfrac{b^2}{4}}$

$\longrightarrow\bf{Apply\:radical\:rule}:\quad \:a=\left(\sqrt{a}\right)^2$

$\bf{\dfrac{1}{4}=\left(\sqrt{\dfrac{1}{4}}\right)^2}$

$\bf{\dfrac{1}{4}b^2-49\:\:=\:=\left(\sqrt{\dfrac{1}{4}}\right)^2b^2-49}$

$\longrightarrow\bf{{Rewrite\:}49\:as\:7^2}$

$=\bf{\left(\sqrt{\dfrac{1}{4}}\right)^2b^2-7^2}$

$\longrightarrow\bf{Apply\:exponent\:rule}:\quad \:a^mb^m=\left(ab\right)^m}$

$\bf{\left(\sqrt{\dfrac{1}{4}}\right)^2b^2=\left(\sqrt{\dfrac{1}{4}}b\right)^2}$

$=\bf{\left(\sqrt{\dfrac{1}{4}}b\right)^2-7^2}$

$\longrightarrow\bf{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\bf{\left(\sqrt{\dfrac{1}{4}}b\right)^2-7^2=\left(\sqrt{\dfrac{1}{4}}b+7\right)\left(\sqrt{\dfrac{1}{4}}b-7\right)}$

$=\bf{\left(\sqrt{\dfrac{1}{4}}b+7\right)\left(\sqrt{\dfrac{1}{4}}b-7\right)}$

$\longrightarrow\bf{Refine}$

$=\bf{\left(\dfrac{b}{2}+7\right)\left(\dfrac{b}{2}-7\right)}$

$\rule{250}{6}$

$\bf{Question\:\:2}$

$\bf{Factorize\:\:a^4-81}$

$\bf{Answer}$

$\bf{Rewrite\:\:}a^4-81\bf{\:as\:}\left(a^2\right)^2-9^2$

$\longrightarrow\bf{How\:\: to \:\:do\:\: that \:\:?}$

$\bf{Apply\:exponent\:rule}:\quad \:a^{bc}=\left(a^b\right)^c$

$\bf{a^4=\left(a^2\right)^2}$

$\bf{Rewrite\:\:}81{\:as\:9^2$

$\longrightarrow\bf{a^4-81\:\:=\:\:\left(a^2\right)^2-9^2}$

$\longrightarrow\bf{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\bf{\left(a^2\right)^2-9^2=\left(a^2+9\right)\left(a^2-9\right)}$

$=\bf{\left(a^2+9\right)\left(a^2-9\right)}$

$\bf{Thankyou !!!!!}$