factorise 16p^2-81q^2
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Answered by
6
16p^2-81q^2
={(4p)^2}-{(9q)^2}
using (a+b)(a-b)=a^2-b^2 - identity
=(4p+9q)(4p-9q)
={(4p)^2}-{(9q)^2}
using (a+b)(a-b)=a^2-b^2 - identity
=(4p+9q)(4p-9q)
Answered by
2
Using identity a^2-b^2=(a+b)(a-b)
16p^2-81q^2=(4p-9q)(4p+9q)
Now 4p-9q=(2√p+3√q)(2√p-3√q)
(From a^2-b^2)
Your answer would be
{(2√p+3√q)(2√p-3√q)}(4p+9q)
16p^2-81q^2=(4p-9q)(4p+9q)
Now 4p-9q=(2√p+3√q)(2√p-3√q)
(From a^2-b^2)
Your answer would be
{(2√p+3√q)(2√p-3√q)}(4p+9q)
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