In this page solving polynomial of degree5 we are going to see how to solve the polynomial which is having degree 5.
Example 1 :
Solve 6 x⁵ + 11 x⁴ - 33 x³ - 33 x² + 11 x + 6
Solution :
The degree of this equation is 5. Therefore we can say there will be 5 roots for this equation.
This is a reciprocal equation of odd degree with like terms. So -1 is one of the root of this equation.
The other roots are given by
6 x⁴ + 5 x³ - 38 x² + 5 x + 6= 0
Dividing the entire equation by x²
6 x⁴/x² + 5 x³/x² - 38 x²/x² + 5 x/x² + 6/x² = 0
6 x² + 5 x - 38 + 5 (1/x) + 6(1/x²) = 0
6 (x² + 1/x²) + 5 (x + 1/x) - 38 = 0 ------ (1)
Let x + 1/x = y
To find the value of x² + 1/x² from this we have to take squares on both sides
(x + 1/x)² = y²
x² + 1/x² + 2 x (1/x) = y²
x² + 1/x² + 2 = y²
x² + 1/x² = y² - 2 solving polynomial of degree5
So we have to plug y² - 2 instead of x² + 1/x²
Let us plug this value in the first equation
6 (y² - 2) + 5 y - 38 = 0
6 y² - 12 + 5 y - 38 = 0
6 y² + 5 y - 12 - 38 = 0
6 y² + 5 y - 50 = 0
6 y² - 15 y + 20 y - 50 = 0
3 y (2y - 5) + 10 (2y - 5) = 0
(3 y + 10) (2y - 5) = 0
3y + 10 = 0
3 y = -10
y = -10/3
2 y - 5 = 0
2 y = 5
y = 5/2
x + 1/x = y
(x² + 1)/x = 5/2
2(x² + 1) = 5 x
2x² + 2 - 5x = 0
2x² - 5x + 2 = 0
2x² - 4x - 1x + 2 = 0
2x (x - 2) -1(x - 2) = 0
(2x - 1) (x - 2) = 0
2x - 1 = 0 x - 2 = 0
2 x = 1 x = 2
x = 1/2
x + 1/x = y
(x² + 1)/x = -10/3
3(x² + 1) = -10x
3x² + 3 = -10 x
3x² + 10 x + 3 = 0
3x² + 9 x + 1 x + 3 = 0
3x (x + 3) + 1(x + 3) = 0
(3x + 1) = 0 (x + 3) = 0
3x = -1 x = -3
x = -1/3
Therefore the 5 roots are x = -1/3,-3,2,1/2,-1
This is the example problem in the topic solving polynomial of degree5. You can try the following sample test to understand this topic much better.
Questions |
Solution |
Question 1 : Solve 6 x⁵ - x⁴ - 43 x³ + 43 x² + x - 6 |
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Question 2 : Solve 8 x⁵ - 22 x⁴ - 55 x³ + 55 x² + 22 x - 8 |
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Question 3 : Solve x⁵ - 5 x⁴ + 9 x³ - 9 x² + 5 x - 1 |
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Question 4 : Solve x⁵ - 5 x³ + 5 x² - 1 |
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Question 5 : Solve 6 x⁵ + 11 x⁴ - 33 x³ - 33 x² + 11 x + 6 |
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