factorise using identity
a) 25a²-4b²+28bc-49c²
b) q²-10q+21
Answers
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Question:
25a²-4b²+28bc-49c²
Step-by-step explanation:
Hope this helps you my friend...
=(5a)²-(2b)²+28bc-(7c)²
=(5a)²-(2b-7c)²
=[5a-(2b-7c)][5a+(2b-7c)]
=(5a-2b+7c)(5a+2b-7c)
Question:
q²-10q+21
Step-by-step explanation:
Hope this helps you my friend...
=q²-(7+3)q+21
=q²-7q-3q+21
=q(q-7)-3(q-7)
=(q-7)(q-3)
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