I really hope there is a math wiz on here that can help me out, I am trying to figure out how to factor hard?

Question

trinomials, can you explain to me how to factor 2a squared+7a+3 Please explain it in simple terms.
also 6x squared-29x+20 your help is greatly needed and appreciated. the reason I am asking for help is because I have a math test tomorrow and don't know how to do this:(

Answer 1

2a^2+7a+3
First you have to set the equation equals 0 then you have

2a^2+7a+3=0
now you find the roots in the equation and get
a=[ -7±sqrt(7^2-4*2*3)]/2*2 =>
a= [-7±sqrt(49-24)]/4=(-7± 5)/4 = -3 or -½
so now you have:
2a^2+7a+3=2(a+3)(a+½)
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The procedure is exactly the same for the next question:

Answer 2

2a^2 + 7a + 3
to factor, you need to understand FOIL, the opposite of factoring. (First, Outside, Inside, Last)

(Ax + B) ( Cx + D) = ACx^2 + ADx + BCx + BD

So you know that in your problem, it has to be (2x + B) (x + D) in order to have 2x^2 to start. We also know that the last number is 3, so B and D have to be a 3 and a 1. We just need to figure out where to put the 2 , 1 , 3, and 1 in order to get the 7a. 2*1 + 3*1 = 5, so that's no good. We need 2*3 + 1 *1 = 7. So that means that the 3 has to be in position D, not B.

So here's the factor:
(2x + 1) (x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3

The next one is
6x^2 - 29 x + 20

Here we have a few more possibilities.
Keeping the same format, we know that if
(Ax + B) (Cx + D), then A * C = 6, and B * D = 20

So, for A, C, we could use 6,1, or 2,3,
and for B,D, we could use 20,1, or 10,2, or 4,5

and we also know that we want AD + BC = 29.
Looking at our number possibilties, I see that 20 * 6 is no good, 20 * 1 + 6 * 1 is no good, 10*2 + 2*3 is no good, or 10*3 + 2*2, so that leaves us with 4 * 6 + 5 * 1, which is 24 + 5 = 29.

So that's the combination of Numbers we need, 6 , 1 for A C and 4,5 for B D

(6x + B) (x + D); we need the 6 to mulitply with the 4, and the 5 with the 1, so we know that D = 4, and B = 5.

(6x + 5) (x + 4) = 6x^2 + 24x + 5x + 20 = 6x^2 +29x +20

Hope this helps!
The key is to look at what the factors are for your Co-efficients, and see how you can use those factors to get your 2 binomials to work.

Answer 3

2a squared+7a+3= 2a^2 +7a +5
If it is factorable it will take this form:
(2a + ?)(a + ?) The 2a*a gives us the 2a^2 we need
The ?*? has to give us the last term 5. The obvious choice is 5*1.
But which question mark is 5 and which is 1? The answer is we have to pick them so we will get a middle term of 7a.
So try (2a+5)(a+1). when we multiply the two inner numbers we get 5a and when we multiply the two outer numbers we get 2a and these add to 7a, So bingo, we got it.

6x^2 - 29x + 20
This one is harder because there are more possibilities. But if you fool around long enough you will come up with:
(6x-5)(x-4) .6x*x gives 6x^2 (5)(-4) gives the +20 and -5x +(-24x gives us our -29x for the middle term.

Hope this helped

Answer 4

2a^2+ 7a+3

In a question like this (ax^2+bx+c), break 'b', so that the sum is equal to b(6 and 1) and the multiple is equal to ac.

2a^2 +6a+a+3

2a(a+3) +1(a+3)

(a+3)(2a+1)

Next Q the same way.

6x^2-24x-5x+20

6x(x-4) - 5 (x-4)

(x-4) (6x-5)

Answer 5

(2a+1)(a+3)

(6x-5)(x-4)