Find the roots of the polynomial equation. x3 - 2x2 + 10x + 136 = 0
Solution:
Given, f(x) = x3 - 2x2 + 10x + 136 = 0
We have to find the roots of the polynomial equation.
Using Descarter’s rule of signs,
f(x) has a 2 change in sign, so there are
1) 2 real roots or
2) a pair of conjugate imaginary roots.
f(-x) = (-x)3 - 2(-x)2 + 10(-x) + 136
f(-x) = x3 - 2x2 - 10x + 136
f(-x) has 1 sign change so there is one real root.
Now use negative values of x, to find the root.
f(-1) = (-1)3 - 2(-1)2 + 10(-1) + 136
= -1 - 2 - 10 + 136
= -13 + 136
f(-1) = 123
f(-2) = (-2)3 - 2(-2)2 + 10(-2) + 136
= -8 - 8 - 20 + 136
= -36 + 136
f(-2) = 100
f(-3) = (-3)3 - 2(-3)2 + 10(-3) + 136
= -27 - 18 - 30 + 136
= -75 + 136
f(-3) = 61
f(-4) = (-4)3 - 2(-4)2 + 10(-4) + 136
= -64 - 32 - 40 + 136
= -136 + 136
f(-4) = 0
Therefore, x = -4 is a root, so (x + 4) is a factor of the given equation.
By long division,
Now, f(x) = (x + 4)(x2 - 6x + 34) = 0
Now find the factor of x2 - 6x + 34 = 0
Testing the discriminant,
D = b2 - 4ac
D = (-6)2 - 4(1)(34)
D = 36 - 136
D = -100
We have a pair of imaginary roots.
Using the quadratic formula,
x = \(\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)
\(x=\frac{-(-6)\pm \sqrt{(-6))^{2}-4(1)(34)}}{2(1)}\\x=\frac{6\pm \sqrt{36-136}}{2}\\x=\frac{6\pm \sqrt{[-100]}}{2}\\x=\frac{6\pm 10i}{2}\\x=3\pm 5i\)
Therefore, the roots are -4, 3 ± 5i.
Find the roots of the polynomial equation. x3 - 2x2 + 10x + 136 = 0
Summary:
The roots of the polynomial equation x3 - 2x2 + 10x + 136 = 0 are -4, 3 ± 5i.