expand a/4-b\2+1 using suitable identities
Answers
answer :
The expanded form is (\frac{a}{4}-\frac{b{2}+1)^2=\frac{a^2+4b^2-4ab+16b+8a+16}{16}( 4a − 2b +1) 2
= 16a 2+4b 2 −4ab+16b+8a+16
Step-by-step explanation:
Given : Expression (\frac{a}{4}-\frac{b}{2}+1)^2( 4a
− 2b +1) 2
To find : The expression value using identity?
Solution :
Expression (\frac{a}{4}-\frac{b}{2}+1)^2( 4a− 2b +1) 2
Using identity,
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)(a+b+c) 2
=a 2 +b 2 +c 2 +2(ab+bc+ca)
Where, a=\frac{a}{4}, b=-\frac{b}{2}, c=1a= 4a ,b=−2b,c=1
Substitute the value,
(\frac{a}{4}-\frac{b}{2}+1)^2=(\frac{a}{4})^2+(-\frac{b}{2})^2+1^2+2((\frac{a}{4})(-\frac{b}{2})+(-\frac{b}{2})(1)+(1)(\frac{a}{4}))( 4a − 2b +1) 2
=( 4a ) 2 +(− 2b ) 2 +1 2 +2(( 4a )(− 2b )+(− 2b )(1)+(1)(
4a ))
(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2}{16}+\frac{b^2}{4})+1+2(\frac{-ab}{8}-\frac{b}{2}+\frac{a}{4})( 4a − 2b+1) 2
= 16a 2 + 4b 2 )+1+2( 8−ab − 2b + 4a )
(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2}{16}+\frac{b^2}{4})+1+2(\frac{-ab-4b+2a}{8})( 4a− 2b +1) 2
= 16a 2 + 4b 2 )+1+2( 8−ab−4b+2a )
(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2}{16}+\frac{b^2}{4})+1+\frac{-ab-4b+2a}{4}( 4a − 2b +1) 2
= 16a 2 + 4b 2 )+1+ 4−ab−4b+2a
(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2+16+4(-ab-4b+2a)}{16}( 4a -2b +1) 2
= 16a 2+4b 2 +16+4(−ab−4b+2a)
(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2+16-4ab+16b+8a}{16}
( 4a − 2b +1) 2
= 16a 2 +4b 2 +16−4ab+16b+8a
(\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2-4ab+16b+8a+16}{16}
( 4a − 2b +1) 2
= 16a 2 +4b 2 −4ab+16b+8a+16
Therefore, The expanded form is (\frac{a}{4}-\frac{b}{2}+1)^2=\frac{a^2+4b^2-4ab+16b+8a+16}{16}( 4a − 2b +1) 2
= 16a 2 +4b 2 −4ab+16b+8a+16