Use the Rational Zero Theorem to find all real zeros

Question

x⁴ + 4x³ − 46x² − 4x + 45 = 0

Answer 1

-9,-1,1,5 are the 4 real zeros, roots, solutions or x intercepts

x^4 + 4x^3 - 46x^2 -4x +45

solve by factoring or graphing

graph the polynomial and the x intercepts = the real zeros

potentially 4 = the polynomial's degree of 4

decartes rule of signs tells us there are

2 positive real zeros and 2 negative real zeros maximum

or could be 0 real zeros and 4 imaginary zeros

or 0 + real zeros and 2 - real zeros and 2 imaginary zeros

or 2 + real zeros and 0 - real zeros and 2 imaginary zeros

as it turns out there are 4 real zeros, 2+, 2- and 0 imaginary zeros

Answer 2

The rational zeros (or roots) must be of the form ±p/q, where p is an integer factor of the constant term and q is an integer factor of the leading coefficient.

x⁴ + 4x³ − 46x² − 4x + 45 = 0

So, p=45 and q=1.

The set of all possible factors is ± { 1, 3, 5, 9, 15, 45 }.

Use either synthetic or polynomial division to determine which of these is an actual factor.

You can also compute f(x) for each of these values; if f(x)=0, then that x is a zero.

Because the degree of the polynomial is 4, there are at most 4 zeros. After you have found 4 roots, you are done.

A graphing calculator can also be used to find the roots.

The roots are { -9, -1, 1, 5}.

Answer 3

x⁴ + 4x³ – 46x² – 4x + 45 = (x⁴ – 46x² + 45) + (4x³ – 4x)

This grouping is motivated by seeing that both groups are readily factored.

(x⁴ – 46x² + 45) = (x² – 45)(x² – 1) and (4x³ – 4x) = 4x(x² – 1)

So now we have

x⁴ + 4x³ – 46x² – 4x + 45 = (x² – 45)(x² – 1) + 4x(x² – 1) = (x² – 45 + 4x)(x² – 1)

x² – 45 + 4x = x² + 4x – 45 = (x + 9)(x – 5) and (x² – 1) = (x + 1)(x – 1)

Giving us the final factored form

x⁴ + 4x³ – 46x² – 4x + 45 = (x + 9)(x – 5)(x + 1)(x – 1)

The real zeros come from solving x + 9 = 0, x – 5 = 0, x + 1 = 0, and x – 1 = 0