Question

Consider the polynomial \(f(x)=a x^{4}+b x^{3}+c x^{2}+d x+e,\) where \(a+b+c+d+e=0 .\) Show that this polynomial is divisible by \(x-1\)

Short answer

The polynomial is divisible by \(x - 1\) because \(f(1) = 0\).

Step-by-step solution

Recall Polynomial Division Theorem

A polynomial is divisible by \(x - 1\) if and only if the polynomial evaluated at 1 is zero. This can be written as \(f(1) = 0\).

Substitute x = 1

Substitute \(x = 1\) into the polynomial \(f(x) = a x^{4} + b x^{3} + c x^{2} + d x + e\). This gives us \(f(1) = a(1)^{4} + b(1)^{3} + c(1)^{2} + d(1) + e = a + b + c + d + e\).

Apply Given Condition

According to the problem, it is given that \(a + b + c + d + e = 0\). Therefore, \(f(1) = a + b + c + d + e = 0\).

Conclusion

Since \(f(1) = 0\), by the Polynomial Division Theorem, the polynomial \(f(x)\) is divisible by \(x - 1\).

Key concepts

polynomial divisibility

In mathematics, a polynomial is said to be divisible by another polynomial if the result of the division is a polynomial with no remainder. For example, if the polynomial \(f(x)\) can be divided by \(g(x)\) without leaving any remainder, then \(g(x)\) is a factor of \(f(x)\). When dealing with polynomial divisibility:

• A polynomial \(P(x)\) is divisible by \(Q(x)\) if \(P(x) = Q(x) \times R(x)\) for some polynomial \(R(x)\).
• In this exercise, the polynomial \(f(x) = a x^{4} + b x^{3} + c x^{2} + d x + e\) is to be checked for divisibility by \(x - 1\).
• To check divisibility, we leverage the polynomial division theorem which simplifies this process.

Therefore, understanding divisibility in polynomials tells us which polynomials serve as factors and helps in factoring more complex polynomials into simpler building blocks. This is particularly useful in solving polynomial equations and in simplifying algebraic expressions.

Polynomial Division Theorem

The Polynomial Division Theorem is a central concept in algebra. It allows us to determine whether a polynomial is divisible by a binomial of the form \(x - c\). The theorem states:

• A polynomial \(f(x)\) is divisible by \(x - c\) if and only if \(f(c) = 0\).
• This means that if you substitute \(c\) into the polynomial and the result is zero, then \(x - c\) is a factor of the polynomial.
• Applying this to the given exercise, we substitute \(x = 1\) into \(f(x)\).

For the polynomial \(f(x) = a x^{4} + b x^{3} + c x^{2} + d x + e\), substituting \(x = 1\) yields: \(f(1) = a (1)^{4} + b (1)^{3} + c (1)^{2} + d (1) + e\), which simplifies to \(f(1) = a + b + c + d + e\). Given that \(a + b + c + d + e = 0\), it follows that \(f(1) = 0\). Hence, by the polynomial division theorem, \(f(x)\) is divisible by \(x - 1\).

evaluating polynomials

Evaluating polynomials involves substituting a specific value for the variable in the polynomial expression and performing the calculations to find the result. The steps are straightforward:

• First, substitute the given value of the variable into the polynomial expression.
• Next, carry out the arithmetic operations following the order of operations: exponentiation, multiplication, and addition.

For the given polynomial \(f(x) = a x^{4} + b x^{3} + c x^{2} + d x + e\), substituting \(x = 1\) gives:

• \(f(1) = a(1)^{4} + b(1)^{3} + c(1)^{2} + d(1) + e\).
• Simplifying, this becomes \(f(1) = a + b + c + d + e\).

Knowing that \(a + b + c + d + e = 0\), it follows that \(f(1) = 0\). This step-by-step evaluation confirms that the polynomial \(f(x)\) is zero at \(x = 1\), supporting the conclusion that the polynomial is divisible by \(x - 1\). Evaluating polynomials at specific points is a fundamental technique used to check polynomial divisibility and to find roots of polynomial equations.