Answer :
[tex]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=\frac{b^2\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]therefore\\x=\frac{-2\pm\sqrt{2^2-4(2)(-3)}}{2(2)}=\frac{-2\pm\sqrt{4+24}}{4}=\frac{-2\pm\sqrt{28}}{4}=\frac{-2\pm\sqrt{4\cdot7}}{4}=\frac{-2\pm2\sqrt7}{4}\\\\=\frac{-1\pm\sqrt7}{2}\\\\so,\ the\ answer:\\\\(x-1)\cdot2\left(x-\frac{-1-\sqrt7}{2}\right)\left(x-\frac{-1+\sqrt7}{2}\right)\\\\=\boxed{(x-1)(2x+1+\sqrt7)\left(x+\frac{1-\sqrt7}{2}\right)}=\boxed{\frac{1}{2}(x-1)(2x+1+\sqrt7)(2x+1-\sqrt7)}[/tex]
[tex]therefore\\x=\frac{-2\pm\sqrt{2^2-4(2)(-3)}}{2(2)}=\frac{-2\pm\sqrt{4+24}}{4}=\frac{-2\pm\sqrt{28}}{4}=\frac{-2\pm\sqrt{4\cdot7}}{4}=\frac{-2\pm2\sqrt7}{4}\\\\=\frac{-1\pm\sqrt7}{2}\\\\so,\ the\ answer:\\\\(x-1)\cdot2\left(x-\frac{-1-\sqrt7}{2}\right)\left(x-\frac{-1+\sqrt7}{2}\right)\\\\=\boxed{(x-1)(2x+1+\sqrt7)\left(x+\frac{1-\sqrt7}{2}\right)}=\boxed{\frac{1}{2}(x-1)(2x+1+\sqrt7)(2x+1-\sqrt7)}[/tex]
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)
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)
