Solve: 0 = x3 + 5x2 + 2x – 8
Solve: 0 = x3 – x2 – 11x – 10
use your graphing calculator or plot it on a graph
Use the Rational Roots Theorem.
Suppose
a x^n + b x^(n-1) + .... e = 0
If x is a rational number of the form p/q with p and q integers that don't have factors in common (i.e. the fraction p/q is simplified as much as possible), then p must divide e and q must divide a.
In case of the equation:
x^3 + 5x^2 + 2x – 8 = 0
this means that x = integer that divides 8. x = 1, x= -2 and x = -4 are solutions.
x^3 – x^2 – 11x – 10 = 0
Assuming again that the roots are rational numbers, it follows from the Rational Roots Theorem that the roots are integers that divide the number 10.
So, the only possible rational roots are:
x = -10, x = -5, x = -2, x = -1, x = 1, x = 2, x = 5, x = 10
Only x = -2 satisfies the equation.
Factoring the polynomial gives:
x^3 – x^2 – 11x – 10 =
(x+2)(x^2 -3x - 5)
The zeroes of the quadratic factor are the other zeroes (you already know that they cannot be rational).
To solve these cubic equations, we can use various methods such as factoring, synthetic division, or using the Rational Root Theorem. In this case, let's try factoring.
1. Solve: 0 = x^3 + 5x^2 + 2x - 8
To solve this equation, we first need to factor it. Let's look for possible integer roots using the Rational Root Theorem. The possible roots are the factors of the constant term (8) divided by the factors of the leading coefficient (1). In this case, the possible roots are ± [1, 2, 4, 8].
By trying these values, we find that x = 1 is a root of the equation. This means (x - 1) is a factor of the equation. So, we can rewrite the equation as:
0 = (x - 1)(x^2 + 6x + 8)
Now we need to solve the quadratic equation (x^2 + 6x + 8 = 0). We can either use factoring or the quadratic formula. In this case, let's use factoring:
0 = (x - 1)(x + 2)(x + 4)
So the solutions to the equation are:
x = 1, x = -2, x = -4
2. Solve: 0 = x^3 - x^2 - 11x - 10
Following the same method, we need to find the factors of the constant term (10) divided by the factors of the leading coefficient (1). Possible roots are ± [1, 2, 5, 10].
By testing these values, we find that x = 1 is a root. This means (x - 1) is a factor of the equation. We can rewrite the equation as:
0 = (x - 1)(x^2 + 2x + 10)
Now we need to solve the quadratic equation (x^2 + 2x + 10 = 0). However, this quadratic equation does not have real solutions because the discriminant (b^2 - 4ac) is negative.
Therefore, the solution to the equation is:
x = 1