SOLUTION: 7a^2+49a+4a+28=0 7a(a+7)+7(a+7)=0 now, (7a+7) (a+7) by solving these, we get a= -1,-7

Question 396661: 7a^2+49a+4a+28=0
7a(a+7)+7(a+7)=0
now,
(7a+7) (a+7)
by solving these, we get
a= -1,-7
Found 3 solutions by CharlesG2, ewatrrr, Parmod:
Answer by CharlesG2(834)   (Show Source): You can put this solution on YOUR website!
"Trying to explain how to split the middle term when solving quadratic equations to my 16 yr old now taking algebra.
Example: 7a(squared)+ 53a + 28
My instinct is to factor it out to:
(7a + 49a +4) + (1a + 4a + 7)
but this is not working when you multiply it out... I just can't quite remember how to split the middle term
Thanks"

7a^2 + 53a + 28 --> you want to factor this, this is not an equation but a trinomial (a 3-term polynomial), if set to zero it would be a quadratic equation
(7a + 4)(a + 7) by FOIL (First Outer Inner Last) this is 7a^2 + 49a + 4a + 28
which adds to the mentioned trinomial

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi, You had the right idea..
7a^2 + 53a + 28 |note the use of the ^2 for 'squared'
(7a + _)(a + _)
What 2 numbers have a 'product of 28' that would work in the blanks?
such that the SUM of the inner product and the outer product = 53a
(7a + 4)(a + 7)
SUM of the inner product(4a) and the outer product(49a) = 53a
7a(a + 7) + 4(a + 7) = 7a^2 + 49a + 4a + 28 = 7a^2 + 53a + 28

Answer by Parmod(1)   (Show Source): You can put this solution on YOUR website!
7a^2+53a+28=0
7a^2+49a+4a+28=0
7a(a+7)+7(a+7)=0
now,
(7a+7) (a+7)
by solving these, we get
a= -1,-7
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