Question

Show that the binomial is a factor of the polynomial. Then factor the function completely. $$ s(x)=x^4+4 x^3-64 x-256 ; x+4 $$

Short answer

Yes, \(x + 4\) is a factor of the function \(s(x)\), and the completely factorized form of \(s(x) = x^4 + 4x^3 - 64x - 256\) is \(s(x) = (x + 4)(x - 4)(x^2 + 4x + 16)\).

Step-by-step solution

Verify that the binomial group is a factor using the Factor Theorem

To verify if \(x + 4\) is a factor, we use the factor theorem. That means we need to substitute \(x = -4\) into the polynomial \(s(x) = x^4 + 4x^3 - 64x - 256\). If we get zero, then \(x + 4\) is a factor. Thus, \(s(-4) = (-4)^4 + 4(-4)^3 - 64(-4) - 256 = 256 - 256 + 256 - 256 = 0\). Therefore, \(x + 4\) is a factor of the function \(s(x)\).

Use Polynomial Long Division

Using polynomial division, divide the function by the factor we just verified, which is \(x + 4\). Thus, we get \(s(x) = (x + 4) (x^3 - 64)\). This is our polynomial divided by the factor.

Factorize the Cubic Polynomial

Further, the expression inside the brackets, \(x^3-64\), is a difference of two cubes. We use the formula \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) to factorize it. Hence, \(x^3 - 64 = (x - 4)(x^2 + 4x + 16)\).

Write the Fully Factorized Function

The fully factorized function is thus \(s(x) = (x + 4)(x - 4)(x^2 + 4x + 16)\).

Key concepts

Factor Theorem

Understanding the Factor Theorem is crucial for factoring polynomials. It is a powerful tool that allows us to test whether a binomial is a factor of a polynomial. The theorem states that for a polynomial function \(f(x)\), if \(f(c)=0\), then \(x-c\) is a factor of \(f(x)\).

In our exercise, we used the Factor Theorem to verify that \(x+4\) is a factor of the polynomial \(s(x)\). We substituted \(x=-4\) into the function and found that \(s(-4)=0\), confirming that indeed \(x+4\) divides \(s(x)\) evenly. It is a straightforward yet effective method to start factoring polynomials, leading the way to reveal other potential factors.

Polynomial Long Division

Polynomial Long Division functions similarly to the long division of numbers. It is a technique used to divide a polynomial by another polynomial of a lower degree. Just like in numerical division, there's a dividend, divisor, quotient, and sometimes a remainder.

When we applied this to our given polynomial \(s(x)\), after verifying \(x+4\) as a factor using the Factor Theorem, we divided \(s(x)\) by \(x+4\) to find the quotient. The absence of a remainder confirms that our divisor \(x+4\) is indeed a factor, and we are left with the quotient \(x^3 - 64\), which simplifies our approach to further factorization.

Difference of Two Cubes

The Difference of Two Cubes is a special case in algebra where we have an expression in the form \(a^3 - b^3\). This type of expression can always be factored into \(a - b)(a^2 + ab + b^2)\).

Our exercise presented the need to factor the quotient \(x^3 - 64\), which is a difference of two cubes where \(a = x\) and \(b = 4\). We applied this formula, obtaining the factors \(x - 4\) and \(x^2 + 4x + 16\). Recognizing patterns like the difference of two cubes simplifies the process of breaking down more complicated polynomials into their prime factors.

Algebraic Expressions

Algebraic expressions are a cornerstone of algebra and involve numbers, variables, and arithmetic operations. The complexity of these expressions can vary, from simple monomials to intricate polynomials like the one in our exercise.

The expression \(s(x)\) that we completely factored in our problem is a polynomial - an algebraic expression consisting of multiple terms that can have various powers. The mastery of factoring algebraic expressions is pivotal in solving equations, simplifying calculations, and understanding the behavior of graphs for these functions.