Question

1. show that the quartic function $f(x)=ax^4 + bx^3 + cx^2 + dx + e$ will always have a rational zero when the numbers $1, -2, 3, 4,$ and $-6$ are randomly assigned to replace $a$ through $e$, and all of the numbers are used.

reflect on your thinking

1. explain how you can use the remainder theorem to locate the possible zeros of a polynomial function.

Explanation:

Question 30

Step 1: Recall Rational Root Theorem

The Rational Root Theorem states that any possible rational zero of a polynomial \( f(x) = a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0\) is of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\). For our quartic function \(f(x)=ax^{4}+bx^{3}+cx^{2}+dx + e\), the possible rational zeros are \(\frac{\text{factors of } e}{\text{factors of } a}\).

Step 2: Analyze the sum of coefficients or specific values

Let's consider \(x = 1\). Then \(f(1)=a(1)^{4}+b(1)^{3}+c(1)^{2}+d(1)+e=a + b + c + d+e\). The numbers assigned to \(a,b,c,d,e\) are \(1,- 2,3,4,-6\). Let's calculate the sum: \(1+( - 2)+3 + 4+( - 6)=(1 + 3+4)+(-2-6)=8 - 8 = 0\).

Step 3: Conclude the rational zero

Since \(f(1)=0\) (because the sum of the coefficients \(a + b + c + d+e = 0\) when we substitute \(x = 1\) into the polynomial), \(x = 1\) is a zero of the polynomial \(f(x)\) for any assignment of the numbers \(1,-2,3,4,-6\) to \(a,b,c,d,e\) (as long as all numbers are used). And \(1\) is a rational number (since it can be written as \(\frac{1}{1}\)). So the quartic function will always have a rational zero (specifically \(x = 1\)).

The Remainder Theorem states that if a polynomial \(f(x)\) is divided by \((x - k)\), the remainder is \(f(k)\). A zero of a polynomial \(f(x)\) is a value \(k\) such that \(f(k)=0\). To locate possible zeros:

1. First, use the Rational Root Theorem to find all possible rational candidates for zeros (values of \(k\)) in the form \(\frac{p}{q}\) (where \(p\) divides the constant term and \(q\) divides the leading coefficient).
2. Then, for each candidate \(k\), evaluate \(f(k)\) (using the Remainder Theorem, which is equivalent to finding the remainder when \(f(x)\) is divided by \((x - k)\)).
3. If \(f(k)=0\), then \(k\) is a zero of the polynomial. By testing these candidate values, we can identify the actual zeros (or narrow down the possible zeros) of the polynomial function.

Answer:

The quartic function \(f(x)\) always has \(x = 1\) as a rational zero because \(f(1)=a + b + c + d+e\) and the sum of \(1,-2,3,4,-6\) is \(0\), so \(f(1) = 0\).

Question 31