Answer to Question #323127 in Algebra for Busi

Question

Let P (x) = 2x^4+15x^3+31x^2+20x+4

(a) Determine whether (x −1) is a factor of P (x). (2)

(b) Find all the possible rational zeros of P (x) by using the Rational Zeros Theorem. (2)

(c) Solve P (x) = 0. (4)

Answer 1

P(x) = 2x⁴ + 15x³ + 31x² + 20x + 4

(a) Determine whether (x −1) is a factor of P (x).

(x − 1) is a factor of P(x) if the sum of the coefficient of P(x) is equal to zero.

Here, the sum of coefficient of P(x) = 2 + 15 + 31 + 20 + 4 = 72, so is not equal to zero.

Hence, (x −1) is not a factor of P (x).

(b) Find all the possible rational zeros of P (x) by using the Rational Zeros Theorem.

The Rational Zero Theorem states that, if the polynomial f(x) has integer coefficients, then every rational zero of f(x) has the form ​p/q​​ where p is a factor of the constant term and q is a factor of the leading coefficient​​.

For P(x) = 2x⁴ + 15x³ + 31x² + 20x + 4 we have:

The constant term is 4; the factors of 4 are p = ±1, ±2, ±4.

The leading coefficient is 2; the factors of 2 are q = ±1, ±2.

​p/q ​​= ​(​Factors of the last) / (Factors of the first) ​​= ±1, ±2, ±4, ±​1/2.

These are the possible rational zeros for the function P(x).

​(c) Solve P (x) = 0.

P(x) = 2x⁴ + 15x³ + 31x² + 20x + 4

Task is to solve: 2x⁴ + 15x³ + 31x² + 20x + 4 = 0.

From (b) we know the possible rational zeros of P(x) are ±1, ±2, ±4, ±​1/2. We can determine which of the possible zeros are actual zeros by substituting these values for x in P(x).

Of those, ±1, ​2, ±4, 1/2 ​​are not zeros of P(x). -2 and -1/2 are the only rational zeros of P(x). This also means that (x + 2) and (x + 1/2) are factors of P(x).

Now we have to find other zeros of P(x).

P(x) = 2x⁴ + 15x³ + 31x² + 20x + 4 after dividing by factor (x + 2) would become P₁(x) = 2x³ + 11x² + 9x + 2.

And P₁(x) = 2x³ + 11x² + 9x + 2 after dividing by factor (x + 1/2) would become P₂(x) = 2x² + 10x + 4.

Lastly, we should solve P₂(x) = 0.

2x² + 10x + 4 = 0

x² + 5x + 2 = 0

In this situation $x=\frac{-b ± \sqrt{\smash[b]{b^2-4ac}}}{2a}$ , where a, b, c are the coefficients of the polynomial.

It means that $x=\frac{-5 ± \sqrt{\smash[b]{5^2-4*1*2}}}{2*1} = \frac{-5 ± \sqrt{\smash[b]{17}}}{2}$.

x = -2 and x = -1/2 are the rational zeros of P(x), other two zeros are $x= \frac{-5 + \sqrt{\smash[b]{17}}}{2}$ and $x= \frac{-5 - \sqrt{\smash[b]{17}}}{2}$.