Factorize : 4x² + 8x + 3?

4x² + 8x + 3
=4x² + 8x + 3
=4x² + 2x + 6x+3
=2x(2x+1)+3(2x+1)
=(2x+1)(2x+3)

( 2x + 3 ) ( 2x + 1 )

Check
2x (2x + 1) + 3 (2x + 1)
4 x ² + 2 x + 6 x + 3
4 x ² + 8 x + 3 (as required)

73.6666666

4x^2 + 8x + 3
=4x^2 + 6x + 2x + 3
= 2x*(2x + 3) + 1*(2x + 3)
= (2x +1) * (2x + 3)
Hence factors r
2x+1 and 2x+3

(2x+3)(2x+1)

Split the middle term:

4x^2 + 8x + 3
= 4x^2 + 2x + 6x + 3
= 2x(2x + 1) + 3(2x + 1)
= (2x + 1)(2x + 3)

This is how you factorize any quadratic polynomial of the form ax^2 + bx + c.
1) Find ac.
2) Find two numbers p and q such that:
(i) pq = ac
(ii) p + q = b
3) Substitute b = p + q in the original polynomial to get:
ax^2 + px + qx + c
= px(qx/c + 1) + c(qx/c + 1)
= (px + c)(qx/c + 1)

4x^2 + 8x + 3
= (2x + 1)(2x + 3)

4x² + 8x + 3= 4x² +8x +12(multiply 4 with 3) now let us see the factors for 12 and 8 (4*3= 12 but 4+3 is not equal to 8,
so let us try another method ,6*2= 12 and 6+2=8 , so we got our factors ) now 4x² +6x+2x+3 (generally u multiply the coefficient of x² with the last number of qd equation to get factors )
now 2x(2x+3)+1(2x+3) = (2x+1)(2x+3)
so answer is x= -1\2or x=-3\2

ax² + bx + c

what is a*c?
12

try to split 12 into numbers that add upto 'b' that is.. 8
mul sum
1 x 12 13
2 x 6 8
3 x 4 7

hence the combination 2 & 6

split the term of x into 2 & 6
4x² + 6x + 2x + 3
factorize:
2x(2x+3) + 1(2x+3)
(2x+1)(2x+3)

incase you cant find such a combination of numbers.. then your factorization wont be perfect anymore.. it would either go into irrational numbers or complex numbers

hope this helps