Respuesta :
The trick to quadratic facotrization is to first check to see if there is a coefficient in front of x^2 that isn't 1. If that is the case you will need to consider two numbers that add to give you the cofficient in front of x and two numbers that multiply to give you a quantity equal to the coefficient in front of x^2 times the number without a variable x. In other words:
[tex]ax^2 + bx + c [/tex]
n+m = b
n*m = a*c
[tex]ax^2 + bx + c [/tex]
n+m = b
n*m = a*c
1) 10x^2+19x+6
a = 10 , b = 19 , c = 6
Start by multiplying a * c = 10 * 6 = 60
Now find two numbers that multiply to 60 and add up to 19...
We could use 15 and 4 because...
15 + 4 = 19
15 * 4 = 60
Now rewrite the expression...
10x^2 + 15x + 4x + 6
Now pull out common factors....
5x ( 2x + 3) + 2 ( 2x + 3 )
The factors of 10x^2+19x+6 are ( 5x + 2) ( 2x + 3)
2) 6x^2+11x+4
a = 6 , b = 11 , c = 4
Start by multiplying a * c = 6 * 4 = 24
Now find two number that multiply to 24 and add up to 11
We could use 8 and 3 ...
8 + 3 = 11
8 * 3 = 24
Now rewrite....
6x^2 + 8x + 3x + 4
Pull out common factors...
2x ( 3x + 4 ) + 1 ( 3x + 4)
The factors of 6x^2+11x+4 are ( 2x + 1 ) ( 3x + 4 )
a = 10 , b = 19 , c = 6
Start by multiplying a * c = 10 * 6 = 60
Now find two numbers that multiply to 60 and add up to 19...
We could use 15 and 4 because...
15 + 4 = 19
15 * 4 = 60
Now rewrite the expression...
10x^2 + 15x + 4x + 6
Now pull out common factors....
5x ( 2x + 3) + 2 ( 2x + 3 )
The factors of 10x^2+19x+6 are ( 5x + 2) ( 2x + 3)
2) 6x^2+11x+4
a = 6 , b = 11 , c = 4
Start by multiplying a * c = 6 * 4 = 24
Now find two number that multiply to 24 and add up to 11
We could use 8 and 3 ...
8 + 3 = 11
8 * 3 = 24
Now rewrite....
6x^2 + 8x + 3x + 4
Pull out common factors...
2x ( 3x + 4 ) + 1 ( 3x + 4)
The factors of 6x^2+11x+4 are ( 2x + 1 ) ( 3x + 4 )