Solve 2p^2 - 3p -2 =0  by factoring?
928398
2p^2 - 3p -2 =0\n(2p+1)(p-2)=0\n2p+1=0\np=-1/2\n\np-2=0\np=2
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
734679
5a^2 + 3a - 2 = 0\n\ncan be factored as (5a - 2) (a + 1) = 0\nThus a = 2/5 and a = -1
417593
no because 8n^2 - 72p^2 can still be factored\n8 is the common factor \n8(n^2-9p^2)\n8(n-3p)(n+3p)
786976
16x^2 + 12x\n\nIn order to solve, you would need an equal sign.  I think maybe you mean finding the zeros?\n\nFirst we would need to factor it.  We could pull an x and a 4 out of each.\n\n4x(4x + 3)\n\nWe get two factors.\n4x\n4x + 3\n\nThen we set each factor equal to 0 to find the zeros of the equation.\n\n4x = 0\nx = 0\n\n4x + 3 = 0\n4x = -3\nx = -3/4\n\nOur zeros are 0 and -3/4.\n\nI hope I helped. :)
255958
Substitute 2 for x and solve for c.\n2^2-3(2)+c=0\nRemember do powers, then multiplication, then subtraction.
220918
2*x^2 - 3*x - 2 = 0 <=> x = -1/2 V x = 2 \n\n==> 2*x^2 - 3*x - 2 <=0 <=> x = [-1/2 , 2]
774014
2^3+2*2^2-3(2)-10=8+8- 6 -10 =0\nTherefore x-2 is a factor of the given polynomial
940135
Use factoring.\nx2-2x-8=0\n(x-4)(x+2)=0\nso\nx-4=0 and x+2=0\nsolve\nx=4 and x=-2
1002101
factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)
259003
Use the quadratic formula.  It works everytime!\nx = (-b plus or minus sprt(b^2 - 4ac))/2\nwhere\nax^2 + bx + c = 0\nSome of these will factor, however not all of them will.
214664
Since the expression p^2-6p-40 is not an equation as written there is no "answer" to the expression. If you are looking for the solution to the equation p^2-6p-40=0, then you can follow what everyone else said. If you want to factor the expression p^2-6p-40 into prime factors, then it will factor to (p-10)(p+4).
796585
1.factor\n(2x+1)(x-4)=0\n2.set it to 0\n2x+1=0\nor\nx-4=0\n3.solve\n{x=-1/2 or x=4} <=answer
188457
you can put 3w^2-6w-24=0 and solve this quadratic equation:\nw= [6+-√(36+4*24*3)]/2*3 = [6+-√(324)]/6=(6+-18)/6\nso w= (6+18)/6=24/6=4\nor w= (6-18)/6= -12/6= -2\nNow you can factor 3w^2 - 6w - 24 = 3(w -4)(w--2)=\n3(w-4)(w+2)
923840
Here's how I would do it:\n\nTo remove the sqrt, you square both sides:\n\nsqrt(p^2 - 3p + 16) = (p+1)(p+1)\np^2 - 3p + 16 = p^2 + 2p + 1\n\nThen it's basic algebra. The p^2 are out of there, and you are left with\n\n-3p + 16 = 2p + 1\n16-1=2p+3p\n15=5p\np=3\n\nSo, let's check:\nsqrt(3^2 - 3(3) + 16) = 3+1\nsqrt(9-9+16)=4\nsqrt 16 =4\n\nThe person who put -5 as one of the answers... it doesn't work:\n\nsqrt ((-5)^2 -3(-5) + 16) = -5+1\nsqrt (25+15+16)= -4\nsqrt 56= -4 This is false, so -5 is not an answer.
865971
p(x)=-3x^2 (x-1) (x+1/3) (x+2)^3\n\n1) what are the roots and multiplicities? \nroots - multiplicity\n0 .............2\n1..............1\n-1/3 .........1\n-2 ............3\nbasically the multiplicity of a root A is the exponent of the factor (x-A)\n.
103674
2x^2 - x = 0  { by adding 1 to both sides }\nx^2 - (1/2)x = 0  { by dividing both sides by 2 }\nx (x - 1/2) = 0 { by factoring }\n\nsolves when either x = 0 or x = 1/2 . QED