Question

the function $f(x) = 5x^3 - 7x^2 - 45x + 63$ has at least one integer root. use the integer root theorem to find that root, then proceed to find all complex roots. (note: roots may be integer, rational, irrational, and/or complex.) answer there are three roots : $-3$, $3$, $\frac{7}{5}$

Explanation:

Step1: Apply Integer Root Theorem

The Integer Root Theorem states that any possible integer root \( p \) of a polynomial \( a_nx^n + \dots + a_1x + a_0 \) divides the constant term \( a_0 \). For \( f(x) = 5x^3 - 7x^2 - 45x + 63 \), the constant term is \( 63 \) and the leading coefficient is \( 5 \). The possible integer roots are the divisors of \( 63 \): \( \pm1, \pm3, \pm7, \pm9, \pm21, \pm63 \).

Test \( x = 3 \): \( f(3)=5(3)^3 - 7(3)^2 - 45(3)+63 = 135 - 63 - 135 + 63 = 0 \). So \( x = 3 \) is a root.

Test \( x=-3 \): \( f(-3)=5(-3)^3 - 7(-3)^2 - 45(-3)+63 = -135 - 63 + 135 + 63 = 0 \). So \( x = -3 \) is a root.

Step2: Factor the Polynomial

Since \( x = 3 \) and \( x = -3 \) are roots, \( (x - 3)(x + 3)=x^2 - 9 \) is a factor. Divide \( f(x) \) by \( x^2 - 9 \):

Using polynomial long division or synthetic division:

\( 5x^3 - 7x^2 - 45x + 63=(x^2 - 9)(5x - 7) \)

Step3: Find All Roots

Set each factor equal to zero:

1. \( x - 3 = 0 \implies x = 3 \)
2. \( x + 3 = 0 \implies x = -3 \)
3. \( 5x - 7 = 0 \implies x=\frac{7}{5} \)

Answer:

The roots of the function \( f(x) = 5x^3 - 7x^2 - 45x + 63 \) are \( -3 \), \( 3 \), and \( \frac{7}{5} \).