1)4+y-
2)5+3-14
plz somebody solve it it's a sum from the factorisation chapter
TPS:
what d you want?
the factorised form of the question
have you missed any a in the second question?
ya sorry!
it's 5+3a-14a^2
1.
4 + y - 14y²
= 4 + 8y - 7y - 14y²
= 4×1 + 2×4 - 7y×1 - 2y×7y
=4( 1+2y) - 7y(1+2y)
=(4-7y)(1+2y)
=(2y+1)(4-7y)
2.
5 + 3a - 14a²
= 5 + 10a - 7a - 14a²
= 5×1 + 5×2a - 7a×1 - 7a×2a
=5(1+2a) - 7a(1+2a)
= (5-7a)(1+2a)
= (2a+1)(5-7a)
4 + y - 14y²
= 4 + 8y - 7y - 14y²
= 4×1 + 2×4 - 7y×1 - 2y×7y
=4( 1+2y) - 7y(1+2y)
=(4-7y)(1+2y)
=(2y+1)(4-7y)
2.
5 + 3a - 14a²
= 5 + 10a - 7a - 14a²
= 5×1 + 5×2a - 7a×1 - 7a×2a
=5(1+2a) - 7a(1+2a)
= (5-7a)(1+2a)
= (2a+1)(5-7a)
Answers
Answered by
2
1.
4 + y - 14y²
= 4 + 8y - 7y - 14y²
= 4×1 + 2×4 - 7y×1 - 2y×7y
=4( 1+2y) - 7y(1+2y)
=(4-7y)(1+2y)
=(2y+1)(4-7y)
2.
5 + 3a - 14a²
= 5 + 10a - 7a - 14a²
= 5×1 + 5×2a - 7a×1 - 7a×2a
=5(1+2a) - 7a(1+2a)
= (5-7a)(1+2a)
= (2a+1)(5-7a)
4 + y - 14y²
= 4 + 8y - 7y - 14y²
= 4×1 + 2×4 - 7y×1 - 2y×7y
=4( 1+2y) - 7y(1+2y)
=(4-7y)(1+2y)
=(2y+1)(4-7y)
2.
5 + 3a - 14a²
= 5 + 10a - 7a - 14a²
= 5×1 + 5×2a - 7a×1 - 7a×2a
=5(1+2a) - 7a(1+2a)
= (5-7a)(1+2a)
= (2a+1)(5-7a)
thank u
you are welcome!!
how did 8y and 7y in the first sum come?
plz tell it
you need to find two numbers such that their sum is b and product is ac.
here b=1 and ac=14*4=56.
so the numbers are 8 and -7.
here b=1 and ac=14*4=56.
so the numbers are 8 and -7.
you need to find those numbers on your own. If you practice more and more, it will be easy for you.
ok thank yuou again
thanks susmita!!
welcome
and u shouldn't say this because this is the reward of your work
Answered by
1
1) -14y²+y+4 = -14y²-7y+8y+4 = -7y(2y+1)+4(2y+1) = (1+2y)(4-7y).
2) -14a²+3a+5 = -14a²-7a+10a+5 = -7a(2a+1)+5(2a+1) = (1+2a)(5-7a).
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