What is the best method to factorise the trinomial 10x^2+16x+6 ???

2 (5x² + 8x + 3)
2 (5x + 3)(x + 1)

10x^2 + 16x + 6
= (10x + 6)(x + 1)

the answer, (5x + 3)(2x + 2)..
i used this method...
5x 3 , cross the number and multiply it..
2x 2 , you will get 16x..
to find the number that i wrote, you must think..what is the number that times a number, will get 10x^2 and 6..of course 5x and 2x. 3 and 2..

but some question..you will need to use 10x and x, 6 and 1, depend to the question request...but the product must be 10x^2 and 6, remember that..

then make the number in bracket, (5x + 3)(2x + 2)..
it formula, dont ask why it must like that...

make sure the coefficient of x is 16, sometime you need to change the position of number before bracket it..

hopefully you understand, try to study it..

1) First, multiply the coefficient of the x^2 term by the constant: 10(6) =60
2) Find factors of 60 that will add to 16(from the middle term 16x).

3) The factors of 60 that add to 16 are +10 and +6.

4)Rewrite the 16x term using +10 +6: 10x^2 +10x + 6x +6

5)Group the four terms two at a time: (10x^2 +10x) + (6x +6)

6)Factor the common factor from each: 10x(x + 1) + 6(x + 1)

7)Factor the common factor in the parenthesis: (x + 1)(10x + 6)

8)Factor a 2 from the 2nd term: 2(x + 1)(5x + 3)

There you go. Hope this helped.

first factor out a two
2(5x^2 + 8x + 3)
2(5x^2 + 5x + 3x +3)
use regrouping method
2( 5x(x + 1) + 3(x + 1)
2(5x +3)(x+1)

If you choose not to factor out the 2 then you could still use regrouping or guess and check

10x^2 + 10x + 6x + 6
10x(x + 1) + 6(x +1)
(10x + 6)(x+1)

Well, there is probably many ways to factor an equation like this, but to make it a lot easier, you should common factor first.

2(5x^2+8x+3)
Then factor that.
2(x+1)(5x+3)
Then multiply by the 2
(2x+2)(5x+3)

(?x +?)(?x+?)
(10x+?)(1x+?) or (5x+?)(2x+?)
The remaining ?'s must be 6,1 or 3,2
You just keep trying all possible combinations until
you find the "fit". If it doesn't work, use the Quadratic
Formula, which works in every case.
As it turns out, (10x+6)(x+1) works.

ax^2 + bx + c
a = 10
b = 16
c = 6

find 2 numbers that multiply to give ac and add to b
ac = 10 * 6 = 60
b = 16

with a little trial and error you find 10 and 6

rewrite equation as 10x^2 + 10x + 6x + 6
now factor in two parts
[10x^2 + 10x] + [6x + 6]
=10x(x + 1) + 6(x + 1)
x + 1 is a common factor now so we get
=(x + 1)(10x + 6)

edit: it seems there is more than one way to factor this
in (10x + 6) you can factor out the 2 leaving 2(5x + 3)(x + 1)
now you can "put" the two back in either set of brackets

(10x + 6)(x + 1) = (5x + 3)(2x + 2)

The answer is:

(10x + 6)(6x + 1)

(5x+3)(2x+2)
The last terms need to multiply to equal 6, while the inner and outer terms must multiply then add to equal 16x. Find the different factors of 10 and 6, and try sets until they work.

seek for the worry-loose high quality/ingredient. a) what's the optimal quantity that the two 6 and 40 8 may well be divided via? that is 6. 6(? +/- ?). Then do 6a/6, and 40 8/6. 6a/6= a, and 40 8/6= 8. 6(a+8). basic peasy. b is the comparable concept, even if that's rather helpful to ingredient out -8, c) comparable element by using fact the different 2, purely with an further variable in touch: 9pq-3p 3p(3q-a million) d) comparable element as c e) do no longer permit the squaring throw you off; the worry-loose ingredient is 'm', or in case you go with for thus, '-m': i will do it the two approaches. -m(m-5n) OR m(-m-5n) mutually as the two are technically acceptable, the 1st is a little extra 'politically' acceptable. f) comparable element, yet unwell help you out with the assumption technique: First, can the 'numerals' be worry-loose factored? hint: sure! 2d, what are the worry-loose variables between the two words? hint: 2 variables! 0.33, what's the backside exponent on each and each of those variables? hint: that could be the quantity outdoors of the brackets. answer: 7xy(4x+y)