Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
[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)
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
