Question

How can I factorise (2x cubed - 5x + 3) ?

Answer 1

2x³ - 5x +3

P = Finding the divisors of the constant 3 : 1 and 3
Q = Finding the divisors of the master coefficient 2 : 1 and 2

Now verify the probably roots: +- p/q
+- 1/1 , +-1/2 , +-3/1 , +-3/2

+-1 , +-1/2 , +-3 , +- 3/2

x=1 is root: 2x³-5x+3 = 2(1)³-5(1)+3 = 0

If x=1 is root, so x-1=0

Divinding by (x-1)

2x³+0x²-5x+3 | x-1
-2x³+2x² 2x² +2x-3
0 +2x²-5x
-2x²+2x
0 -3x +3
3x - 3
0 0

So,
2x³-5x+3 = (x-1)(2x²+2x-3)

(x-1)(2x²+2x-3)

Answer 2

`2x^3-5x+3=2x^3-2x-3x+3\\\\=2x(x^2-1)-3(x-1)\\\\=2x\underbrace{(x^2-1^2)}_(use\ (*))-3(x-1)\ \ \ |(*)\ a^2-b^2=(a-b)(a+b)\\\\=2x\underbrace{(x-1)}_((**))(x+1)-3\underbrace{(x-1)}_((**))\\\\=(x-1)[2x(x+1)-3]\\\\=(x-1)\underbrace{(2x^2+2x-3)}_((***))\\\\(***)\ 2x^2+2x-3\to a=2;\ b=2;\ c=-3\\\\x=(b^2\pm√(b^2-4ac))/(2a)`


`therefore\\x=(-2\pm√(2^2-4(2)(-3)))/(2(2))=(-2\pm√(4+24))/(4)=(-2\pm√(28))/(4)=(-2\pm√(4\cdot7))/(4)=(-2\pm2\sqrt7)/(4)\\\\=(-1\pm\sqrt7)/(2)\\\\so,\ the\ answer:\\\\(x-1)\cdot2\left(x-(-1-\sqrt7)/(2)\right)\left(x-(-1+\sqrt7)/(2)\right)\\\\=\boxed{(x-1)(2x+1+\sqrt7)\left(x+(1-\sqrt7)/(2)\right)}=\boxed{(1)/(2)(x-1)(2x+1+\sqrt7)(2x+1-\sqrt7)}`