how do you factor 2c^2 + 7c - 4?

Answer 1

(2c-1)(c+4)

Answer 2

first in order to get 2c^2 you know that one factor will start with 2c and the other with c. Also since you have a plus and a minus in your polynomial you will have both in your factors as well. to get the last part you need to think of all the factors of 4 ( 1x4 or 2x2) now one of these sets will be the last half of your factors. if it doesn't look obvious try playing around with it just don't forget about the 2c.
(2c-1)(c+4)

Answer 3

I wanna learn u a method:when the coeffient of the squared letter isn't a complete square you can multipy (a) by (c) and then factor and after all devide them again to the coeffient of the squared letter(I mean a!) like in the one that u've written:
2*-4=-8
(2c+8)(2c-1)/2

Answer 4

This is not factorable, but you can use the quadratic formula to find its roots. I made a mistake. It is factorable try (2c-1)times(c+4)

Answer 5

2c^2 + 7c - 4
2c^2 -c + 8c -4
c(2c-1) + 4(2c -1)
(2c-1)(c+4)

Answer 6

(2c-1)(c+4)

Answer 7

(2c-1)(c+4)

Answer 8

(c+4)(2c-1)

Answer 9

I would just use the quadratic formula on it.

[-b +/- sqrt(b^2-4*a*c)] / 2a

where in your case a = 2, b = 7, and c = -4

by solving the + and - side of this equation you get your two roots.

Answer 10

I have used x instead of c. You can change it.
This equation is of form ax2+bx+c
a = 2 b = 7 c = -4
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-7 +/-sqrt(7^2-4(2)(-4)]/(2)(2)
discriminant is b^2-4ac =81
x=[-7 +sqrt(81)] / (2)(2)
x=[-7 -sqrt(81)] / (2)(2)
x=[-7+9] / 4
x=[-7-9] / 4
The roots are 0.5 and -4