What is the root of the polynomial equation x(x - 2)(x + 3)=18?
Use a graphing calculator and a system of equations.

Solution:

Given polynomial equation x(x - 2)(x + 3)=18

Using the Rational Zeros Theorem, which states that, if the polynomial f(x) = a_nxⁿ + a_n−1x^n-1 +...+ a1x + a0 has integer coefficients, then every rational zero of f(x) has the form p/q where p is a factor of the constant term a0 and q is a factor of the leading coefficient a_n.

Here p : ±1, ±2, ±3, ±6, ±9, ±18 which are all factors of constant term -18.

q: ±1

Given x(x-2)(x+3)=18 ⇒ x³ + x² -6x -18 =0

f(x) = x³ + x² -6x -18

By inspection,

f(1) = 1+1 -6 -18 = -22 ≠ 0

f(-1) = -1+1+6-18 = -12 ≠ 0

f(2) = 8 +4-12-18 = -18 ≠ 0

f(-2) = -8 +4+12-18 = -10 ≠ 0

f(3) = 27 +9-18-18 = 0

By synthetic division:

The remainder is 0

The quotient is a quadratic polynomial.

x² +4x +6 = 0

Discriminant of the quadratic equation, Δ = b² - 4ac

Here, b =4, a = 1, c = 6

∴ Δ = b² - 4ac = 16 - (4×1×6) = -8

 Δ is negative. Hence no real solution exists.

The real root is 3 and the other complex roots are -2 ± √2i.

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What is the root of the polynomial equation x(x - 2)(x + 3)=18? Use a graphing calculator and a system of equations.

Summary:

For the given polynomial equation x(x - 2)(x + 3)=18, The real root is 3 and the other complex roots are -2 ± √2i