Solve the following polynomial equations. 2x(4x-7)(x+3)^2=0 & (x+3)(x+2)(x-2)=-12?

1) 2x(4x-7)(x+3)^2=0
=>2x(4x-7)(x+3)(x+3)=0
=>2x=0 or 4x-7=0 or x+3=0 or x+3=0
=>2x=0 or 4x=7 or x= -3 or x= -3
=>x=0/2, 7/4, -3
:.x= -3, 0, 1.75 (or 1 3/4)

2) (x+3)(x+2)(x-2)= -12
=>(x+3)(x^2 - 2^2)= -12
=>(x+3)(x^2 -4)= -12
=>x(x^2 -4)+3(x^2 -4)= -12
=>x^3 - 4x + 3x^2 - 12 = -12
=>x^3+3x^2-4x=0
=>x(x^2+3x-4)=0
=>x(x^2+4x-x-4)=0
=>x(x+4)(x-1)=0
=>x=0 or x+4=0 or x-1=0
:.x= -4 , 0 , 1

in case you element out (2x - 3), then all will become sparkling. often, once you get a cubic, attempt some elementary factors, first. Like, in case you notice that putting in a million can provide an answer under 0, yet 2 can provide an answer greater desirable than 0, then hunt for a element between there. If that would not artwork, there are primary techniques for fixing a cubic - even a widespread formulation. those could be stumbled on on Wikipedia.

1)
2x(4x - 7)(x + 3)^2 = 0

2x = 0
x = 0/2
x = 0

4x - 7 = 0
4x = 7
x = 7/4 (1.75)

x + 3 = 0
x = -3

∴ x = -3 , 0 , 7/4 (1.75)

= = = = = = = =

2)
(x + 3)(x + 2)(x - 2) = -12
(x + 3)(x^2 - 4) + 12 = 0
x^3 + 3x^2 - 4x + 12 = 0
(x^3 + 3x^2) - (4x - 12) = 0
x^2(x + 3) - 4(x - 3) = 0
(x - 3)(x^2 - 4) = 0
(x - 3)(x + 2)(x - 2) = 0

x - 3 = 0
x = 3

x + 2 = 0
x = -2

x - 2 = 0
x = 2

∴ x = 3 , ±2

Have you ever heard of Please Excuse My Dear Aunt Sally? or P.E.M.D.A.S? Well it stands for Parenthesis, Exponents, Multiplication, Division, Addition, and Subtraction. In case you aren't familiar with these, they are the order of operations. You can apply them to this situation =]

Parenthesis: nothing to do
Exponents: aha! we have one here =] so that means...

Step 1: Use the Distribution Property with the exponent and this will remove the second set of parenthesis. It should look like this:
2x(4x-7)x^2+3^2=0 this then simplifies to 2x(4x-7)x^2+9=0

Parenthesis: nothing
Exponents: nothing
Multiplication: ta da!

Step 2: Use the Distribution Property with 2x and the remaining set of parenthesis. It should look like this:
(8x^2-14x)(x^2)+9=0

Parenthesis: nothing
Exponents: nothing
Multiplication: yup

Step 3: Yet again, use the Distributive Property with x^2 and the other set of parenthesis. It should look like this:
8x^4-14x^2+9=0

Now, to solve for x you must isolate it on one side of the equation by itself and you will have the answer.

Step 4: Subtract nine from both sides. It should look like this:
8x^4-14x^2=-9

I think you should know what to do from here? i am tired so i need some sleep, but good luck with the rest! =D

1) 2x(4x-7)(x+3)^2=0
2x=0 4x-7=0 x+3=0 x+3=0
x=0 4x=7 x=-3 x=-3
..........x=7/4

x=0,7/4,-3,-3 answer//


2) (x+3)(x+2)(x-2)=-12
(x^2+5x+6)(x-2)=-12
x^3+5x^2+6x-2x^2-10x-12=0
x^3+3x^2-4x-12=-12
x^3+3x^2-4x-12+12=0
x^3+3x^2-4x=0
x(x^2+3x-4)=0
x=0 (x+4)(x-1)=0
x=0 x+4=0 x-1=0
x=0 x=-4 x=1

x=0,-4,1 answer//