from a handpicked tutor in LIVE 1-to-1 classes

# Find the roots of the polynomial equation. x^{3} - 2x^{2} + 10x + 136 = 0

**Solution:**

Given, f(x) = x^{3} - 2x^{2} + 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) = x^{3} - 2x^{2} - 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)(x^{2 }- 6x + 34) = 0

Now find the factor of x^{2} - 6x + 34 = 0

Testing the discriminant,

D = b^{2 }- 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. x^{3} - 2x^{2} + 10x + 136 = 0

**Summary:**

The roots of the polynomial equation x^{3} - 2x^{2} + 10x + 136 = 0 are -4, 3 ± 5i.

visual curriculum