please factorize two expressions above
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2nd answer:: 6a^2+2b^2+3c^2+7ab+7bc+11ac
=2b^2+7b(a+c)+(6a^2+11ac+3c^2)
=2b^2+7b(a+c)+(6a^2+9ac+2ac+3c^2)
=2b^2+7b(a+c)+{3a(2a+3c)+c(2a+3c)}
=2b^2+7b(a+c)+(2a+3c)(3a+c)
=2b^2+{(4a+6c)+(3a+c)}b+(2a+3c)(3a+c)
=2b^2+(4a+6c)b+(3a+c)b+(2a+3c)(3a+c)
=2b(b+2a+3c)+(3a+c)(b+2a+3c)
=(b+2a+3c)(2b+3a+c)
and 1st answer is attached as pic .
=2b^2+7b(a+c)+(6a^2+11ac+3c^2)
=2b^2+7b(a+c)+(6a^2+9ac+2ac+3c^2)
=2b^2+7b(a+c)+{3a(2a+3c)+c(2a+3c)}
=2b^2+7b(a+c)+(2a+3c)(3a+c)
=2b^2+{(4a+6c)+(3a+c)}b+(2a+3c)(3a+c)
=2b^2+(4a+6c)b+(3a+c)b+(2a+3c)(3a+c)
=2b(b+2a+3c)+(3a+c)(b+2a+3c)
=(b+2a+3c)(2b+3a+c)
and 1st answer is attached as pic .
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Anonymous:
I have a doubt in a area can you clear that one ?
of which figure?
where is your ans.
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wclm
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hey mate I have solved your question. hope it will help you.
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I too did the same thing right but in a short cut
ok
Good work
the ansqer will be same frim any method
ok
yeah
thanks to both of you
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