Question

Factor the following trinomial:

20a^2 + 51ab + 28b^2

Answer 1

@campbell_st

Answer 2

I don't have the patience for this one lol

Answer 3

I know the factors for 20: 20x1,5x4,10x2 & 28 are: 28x1,7x4,14x2

Answer 4

ok...20 x 28 = 480

find the factors that add to 51...

so its 35 and 16

so again

(20a + 35b)(20a + 16b)
----------------------
20

Answer 5

same as before... remove the common factors from each binomial... and they cancel with the denominator

Answer 6

I don't understand how to do it like this...

Answer 7

ok.... in this question I ignored b as both binomials will have a term in b


but basically

for a quadratic

\[ax^2 + bx + c \]

start by multiplying a and c in your question is 20 and 28 = 560

find the factors that add to the coefficient of the middle term (b) in the standard form above.

in your question I needed 2 factors that add to 51.... hence 35 andd 16
( I used a factor tree to find them)

now you write the equation as

(ax + factor 1)(ax + factor 2)
--------------------------
a

in your question it was

(20a + 35b)(20a + 16b)
---------------------
20

almost there... factor the binomials... and if done correctly it always cancels
with the denominator...a

so

5(4a + 7b)4(5a + 4b)
------------------
20

or

5*4(4a+7b)(5a + 4b)
-------------------
20


can you can now cancel the common factor leaving

(4a + 7b)(5a + 4b)

and it works every time.

Answer 8

Ooooooh ok ! Thanks!

Answer 9

lol... I'd bet your teacher hasn't seen it... good luck