which of the following are roots of the polynomial function below f(x)=2x^3-x^2-9x+6
f(x) = 2x³ - x² - 9x + 6
0 = 2x³ - x² - 9x + 6
0 = 2x³ + 3x² - 4x² - 3x - 6x + 6
0 = 2x³ + 3x² - 3x - 4x² - 6x + 6
0 = x(2x²) + x(3x) - x(3) - 2(2x²) - 2(3x) + 2(3)
0 = x(2x² + 3x - 3) - 2(2x² + 3x - 3)
0 = (x - 2)(2x² + 3x - 3)
0 = x - 2 or 0 = 2x² + 3x - 3
+ 2 + 2 x = -3 ± √(3)² - 4(2)(-3)
2 = x 2(2)
x = -3 ± √9 - 4(-6)
4
x = -3 ± √9 + 24
4
x = -3 ± √33
4
x ≈ -3 ± 5.74
4
x ≈ -3 + 5.74 or x ≈ -3 - 5.74
4 4
x ≈ 2.74 or x ≈ -8.74
4 4
x ≈ 0.685 or x ≈ -2.185
Answer:
The roots of the given polynomial are
\(x=2,\frac{-3+\sqrt{33}}{4},\frac{-3-\sqrt{33}}{4}\)
Step-by-step explanation:
Given : Polynomial \(f(x)=2x^3-x^2-9x+6\)
To find : The roots of the given polynomial?
Solution :
Using hit and trial method one of the root of the polynomial is 2 as 2 satisfy the given polynomial equating zero.
Let x=2 substitute in polynomial,
\(f(2)=2(2)^3-2^2-9(2)+6\)
\(f(2)=2(8)-4-18+6\)
\(f(2)=16-4-18+6\)
\(f(2)=22-22\)
\(f(2)=0\)
So, One the root is 2.
Now, We divide the \(f(x)=2x^3-x^2-9x+6\) by (x-2) to get the remainder in quadratic form by long division method.
The remainder is \(2x^2+3x-3\)
Now, we factor the quadratic equation by using quadratic formula,
General form - \(ax^2+bx+c=0\)
\(D=b^2-4ac\)
Solution is \(x=\frac{-b\pm\sqrt{D}}{2a}\)
Equation is \(2x^2+3x-3\)
where, a=2 , b=3, c=-3
\(D=b^2-4ac\)
\(D=(3)^2-4(2)(-3)\)
\(D=9+24\)
\(D=33\)
Solution is \(x=\frac{-b\pm\sqrt{D}}{2a}\)
\(x=\frac{-(3)\pm\sqrt{33}}{2(2)}\)
\(x=\frac{-3\pm\sqrt{33}}{4}\)
\(x=\frac{-3+\sqrt{33}}{4},\frac{-3-\sqrt{33}}{4}\)
Therefore, The roots of the given polynomial are
\(x=2,\frac{-3+\sqrt{33}}{4},\frac{-3-\sqrt{33}}{4}\)
