If you mean: (3x+1) and (2x+3) are factors of 6xsquared+11x+3 then yes they are
6x2 + 11x + 3 = 6x2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1)
6x2-5x-25 = (3x+5)(2x-5) when factored
6x2 + 11x - 10 = (6x2 + 15x) - (4x + 10) = 3x(2x + 5) - 2(2x + 5) = (3x - 2)(2x + 5). Note, first, that the three co-efficients of the original quadratic expression are 6, 11, and -10. We need two numbers whose sum is 11 and whose product is (6)(-10) = -60. These two numbers turn out to be 15 and -4. Thus, we replace the middle term, 11x with 15x - 4x and proceed with the factorisation in the usual way.
6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y)
(2x - 5) is a binomial factor
6x2 + 11x + 3 = 6x2 + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1)
6x2-5x-25 = (3x+5)(2x-5) when factored
The expression may be : 6x2 + 17x + 12 This factors as, 6x2 + 17x + 12 = (3x + 4)(2x + 3) Or, the expression could be : 6x2 - 17x + 12 This factors as, 6x2 - 17x + 12 = (3x - 4)(2x - 3)
6x2 + 11x - 10 = (6x2 + 15x) - (4x + 10) = 3x(2x + 5) - 2(2x + 5) = (3x - 2)(2x + 5). Note, first, that the three co-efficients of the original quadratic expression are 6, 11, and -10. We need two numbers whose sum is 11 and whose product is (6)(-10) = -60. These two numbers turn out to be 15 and -4. Thus, we replace the middle term, 11x with 15x - 4x and proceed with the factorisation in the usual way.
You have to experiment with different factors of 2, and different factors of 15. In this case, 2x2 - 11x + 15 = (2x - 5) (x - 3)
6x2 + 14x -12 = (3x - 2) (2x + 6). There they are.
6x2=12 2x-28=?
6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y) 6x2+ 35xy + 11y2=6x2+ 33xy + 2xy + 11y2= 3x(2x + 11y) + y(2x + 11y)= (2x + 11y)*(3x + y)
If that's + 6, that factors to (3x - 2)(2x - 3)
11x
No. It factors to (11x - 13)(2x - 3)
(2x - 5) is a binomial factor