solve it x^3-3x^2-16x+48=0?

Answer 1

x^3-3x^2-16x+48=0
By trial & Error method putting x=3 the equation becomes o,
so,
x^2(x - 3) - 16(x - 3) = 0
(x^2 - 16)(x - 3) = 0
(x+4)(x-4)(x-3) = 0
x=-4, +4 or 3

Answer 2

i will kick you off with the first few, i have in basic terms get a couple of minutes. a million) 3x^2 + 7x = -2 so 3x^2+7x+2=0 now in basic terms 3 and a million divide 3 because this is best so the brackets look like (3x )(x )=0 now in basic terms 2 and a million divide 2 because this is best and three*2 + a million*a million = 7 so the brackets are (3x+a million)(x+2)=0 now both (3x+a million)=0 or (x+2)=0 so x= -a million/3 or x=-2 2) x^2-40 9=0 40 9 is 7 squared and also you want the elements of x to cancel so (x+7)(x-7)=0 so x = -7 or 7 3)x^2+5x+4=0 so favor 2 factors of four that upload as a lot as 5 hence (x+4)(x+a million)=0 so x=-4 or -a million run out of time yet i have defined a thanks to do it so that you would possibly want to be in a position to end something

Answer 3

x^3-3x^2-16x+48=0
First we factorise
x^2(x-3)-16(x-3)=0
(x^2-16)(x-3)=0
Then we solve the two equation we have derived
x^2-16=0 and x-3=0
x^2=16 and x=3
On the first equation square root both sides of the equation, therefore it will be
x=4

Finally
The solution for x=4 or x=3

Answer 4

x^3 - 3x^2 - 16x + 48
(x^3 - 3x^2) + (-16x + 48)
x^2(x - 3) - 16(x - 3)
(x^2 - 16)(x - 3)
(x - 4)(x + 4)(x - 3)
x = 4, -4, or 3

Answer 5

x = -4, 3 or 4

Answer 6

x^3-3x^2-16x+48=0
=> (x-3)(x^2-16)=0
=> (x-3)(x-4)(x+4)=0
So x=3,4,-4

Answer 7

x^3-3x^2-16x+48=0
x^2(x - 3) - 16(x - 3) = 0
(x^2 - 16)(x - 3) = 0
(x+4)(x-4)(x-3) = 0
x=-4, +4 or 3

Th

Answer 8

x^2(x-3)-16(x-3)=0
(x-3)(x^2-16)=0
x=3.....x=4

Answer 9

x^2(x-3)-16(x-3)=0
(x^2-16)(x-3)=0
(x+4)(x-4)(x-3)=0
x=-4
x=4
x=3

Answer 10

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