solve by factoring?

1)
4x^3 - 9x = 0
x(4x^2 - 9) = 0
x(2x + 3)(2x - 3) = 0

x = 0

2x + 3 = 0
2x = -3
x = -3/2 (-1.5)

2x - 3 = 0
2x = 3
x = 3/2 (1.5)

∴ x = 0 , ±3/2 (±1.5)

= = = = = = = =

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

x = 0

2x - 1 = 0
2x = 1
x = 1/2 (0.5)

x + 4 = 0
x = -4

∴ x = 0 , 1/2 (0.5) , -4

= = = = = = = =

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

x + 2 = 0
x = -2

x - 2 = 0
x = 2

x - 3 = 0
x = 3

∴ x = ±2 , 3

= = = = = = = =

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

x + 1 = 0
x = -1

x - 1 = 0
x = 1

2x - 1 = 0
2x = 1
x = 1/2 (0.5)

∴ x = ±1 , 1/2 (0.5)

1. x([2x]^2 - [3^2]
= x(2x + 3)(2x - 3)
2. x(2x^2 + 7x - 4)
x (2x - 1)(x + 4)
3. There is a procedure to be followed for factoring a polynomial with a cube of the unknown. Procedure gets simplified if the coefficient of square term(of x^2) is zero. So,
put x=z+c. Then the eqn. becomes
(z+c)^3-3(z+c)^2-4(z+c)+12
= z^3+3z^2c+3zc^2+c^3-3z^2-6zc-3c^2-4z-4c+12
=z^3+z^2(3c-3)+z(3c^2-6c-4)+(c^3-3c^2-4c+12),
put c = 1 to make the coeff. of 'z^2' zero, then it becomes
z^3 +z(3-6-4)+(1-3-4+12),
z^3 - 9z + 6 = 0 needs to be solved, in which the coeffs. are
a' =6, a"= -7.
The procedure is -
let q = a"/3 = -7/3; r = a'/2 = 3.
The radical {q^3+r^2}=R:
If {q^3+r^2} >0, it gives '1 real root & a pair of complex roots'
=0 gives 'all real roots, with a pair of equal roots',
< 0 gives all 'real roots'. so,
Since R = {-100/27} < 0 & it gives all real roots.
sqrt{-(343/27) + 9} = [1/3 sqrt3] sqrt{-100} = j10[1/(3 sqrt3)] that is imaginary.
Here we introduce a conjugate pair
s1 = cuberoot[r +sqrt{q^3+ r^2}],
s2 = cuberoot[r - sqrt{q^3+ r^2}].
So,
s1=cuberoot[3 + j(10/3)/sqrt3]= (1/sqrt3)cuberoot[9 sqrt3+j10]
s2=(1/sqrt3)cuberoot[9 sqrt3- j10]
since cuberoot (3 sqrt3)= sqrt3.
Here it is the cuberoot of a complex number in rectangular coordinates. Convert this to polar coordinates that give 'magnitude'/'angle'. Take the cube root of 'magn.' and divide 'angle' by 3. Magn happens to be
cuberoot[sqrt343]=sqrt[cuberoot(7^3)]=7^(1/2)=sqrt7.
The cuberoot finallty is (sqrt27/2 [+/-] 1/2)
s1, s2=(1/sqrt3)[(sqrt27/2)[+/-] j(1/2)]
=(3/2)[+/-] j(1/2 sqrt3)
Now the 3 roots of third degree polynomial in 'z' are
z1= sum of s1= s1+s2 and the imaginary part goes.
z2, z3 = {-sum [+/-] j sqrt3(difference)}/2. So,
z1 = 3
z2, z3 = {-3/2 [+/-] (j^2)(1/4)}
= - {(3/2) [+/-] (1/4)}
= - (7/4), -(5/4).
Now to get 'x' back,
x =z+1.
x1 = z1+1 = 4
x2 = z2+1 = 1 -(7/4) = -(3/4)
x3 = z3+1 = 1 -(5/4) = -(1/4)
Hence the factorized polynomial is
(x-4)(x+ 3/4)(x+ 1/4)
=(x-4)(4x+ 3)(4x+ 1) /16.
4. 2x^3-x^2-2x+1=0.
2x(x^2-1)-(x^2-1)=0
(x^2-1)(2x-1).

1 ). .......4x^3 - 9x = 0
x(4x² - 9) = 0
============
Ans: : x = 0
============
( 4x² - 9) = 0
4 x² = 9
x² = 9/4
x² = (3/2)(3/2)
=============================
Ans: :x = 3/2 : x= -3/2: x =0
=============================

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

2x³ + 7x² - 4x =0
x(2x² + 7x - 4) = 0
===============
Ans: : x = 0
===============
( 2x² + 7x -4) = 0
Quadratic equation:
a = 2: b = 7: c = -4
x = [ -b ±√ (b)² - 4ac] /2a
x = [-(7) ±√ (7)² - 4(2)(-4) ] /2(2)
x = [ -7 ±√ 49 + 32 ] /4
x = [-7 ±√ 81 ] /4
Take positive sign
x = [ - 7 + 9 ] /4
x = 2 / 4
=================
Ans: : x = 1/2
=================
Take negative sign
x = [ - 7 - 9 ] /4
x = - 16 / 4
=============================
Ans: : x = -4 : x = 1/2 : x = 0
=============================

3). ......x^3-3x^2-4x+12=0
x³ - 3x² -4x + 12 = 0
x³ - 3x² - 4x = -12
x(x² - 3x -4) = -12
================
Ans: : x = -12
================

(x² - 3x -4) = -12
x(x - 3) = 4 -12
x (x - 3) = -8
================
Ans: :x = -8
================
(x- 3) = -8
x = -8 + 3
===========================
Ans: : x = -5: x = -8: x = -12
===========================

4) .......x^2(2x-1)=2x-1
x²(2x - 1) = 2x -1
x² (2x - 1) - (2x - 1) = 0
(2x - 1) ( x² - 1) = 0
(2x -1) = 0
2x = 1
================
Ans: : x = 1/2
================
x² - 1 = 0
x² = 1
===========================
Ans: : x = -1 & x = 1: x = 1/2
===========================

4x³ - 9x = 0
x(4x² - 9) = 0
x(2x - 3)(2x + 3) = 0
x = 0 or (2x - 3)= 0 or (2x + 3) = 0
x = 0 or x = 3/2 or 2x = – 3/2

2x³ + 7x²- 4x = 0
x (2x² + 7x - 4) = 0
x [2x² – x + 8x - 4] = 0
x [x(2x – 1) + 4(2x - 1] = 0
x (2x – 1)(x + 4) = 0
x = 0 or (2x – 1)= 0 or (x + 4) = 0
x = 0 or x = 1/2 or x = – 4

x³ - 3x²- 4x + 12 = 0
x²(x - 3) - 4(x – 3) = 0
(x² - 4)(x – 3) = 0
(x² - 4) = 0 or (x – 3) = 0
x = ± 2 or x = 3

x²(2x-1) = (2x-1) ---- divide by (2x – 1)
x² = 1
x = ± 1
---------

Did Paldin forget to solve and the whole of the fourth?
As far as factorisation, he is correct till thire.
here is the complete.

1)x(2x+3)(2x-3) = 0 so x = 0,3/2, -3/2

2)2x(x^2+7x-2)=0. So x =0 or [-7 + or - Rt65]/2.

3)(x-3)(x+2)(x-2)=0 so x = 3 or + or - 2.

4)Either 2x - 1 = 0 or x^2 =1 so x = 1/2 or 1 or -1.

for the first one, factor out an x, then factor the difference of squares

with the second, also start by factoring out an x

with the third, factor by grouping

***( I didn't forget anything, just trying to get things rolling and let him finish on his own)

4x^3 - 9x = 0
x(x^2-9)=0
x((x-3)(x+3))=0
so x=0 or x=3 or x=-3

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