Assuming the equation is 2b2+5b-35=0, the quadratic formula is one way to solve this equation. It can't be factored easily. Do not confuse the "b" variable in the formula given with the "b" in the related link. The values of a, b, & c are as follows to be used in the quadratic formula: a=2, b=5, c= -35. Placing these values in the quadratic formula gives: (-5+/- sqrt(52-4*2*(-35)))/(2*2). This yields variable b = (-5+/- sqrt(305))/4 or (-5+17.46425)/4 & (-5-17.46425)/4 = 3.11606 & -5.61606 which are the two roots of the given equation.
24a + b2 + 3a + 2b2= 27a + 3b2
We can combine the like terms. So the b2 can be combined with the 2b2 to give 3b2. Likewise the 3b plus the -5b gives -2b.Therefore, b2 + 3b - 5b + 2b2 = 3b2 - 2b.
The Answer to this question is 9 (2B+4) and (1B+3)
3a(b+c)+2b(b+c)
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
24a + b2 + 3a + 2b2= 27a + 3b2
(a2+2b2-2ab)(a2+2b2+2ab)
We can combine the like terms. So the b2 can be combined with the 2b2 to give 3b2. Likewise the 3b plus the -5b gives -2b.Therefore, b2 + 3b - 5b + 2b2 = 3b2 - 2b.
= 4a2 + 2ab 2b2
2b2 + 8 para b = -3
The Answer to this question is 9 (2B+4) and (1B+3)
3a(b+c)+2b(b+c)
b2 + b2 = 2b2 (when terms are alike, just add them up)
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
3ab + 3ac + 2b2 + 2bc = 3a(b + c) + 2b(b + c) = (3a + 2b)(b + c)
do you mean 12a3 - 5a2b - 2b2 ?? = 12a3 + 8a2b - 3a2b - 2b2 then group 1st 2 terms and last 2 and use GCF = (12a3 + 8a2b) + ( - 3a2b - 2b2) = 4a2(3a + 2b) + (-b)(3a2 +2b) not much help, but that's all folks
6b2c3