Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly 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)
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
