Problem 17 - Precalculus

Question

Express each polynomial in the form \(a_{n}\left(x-r_{1}\right)\left(x-r_{2}\right) \cdots\left(x-r_{n}\right)\). $$x^{5}-7 x^{3}-18 x$$

Step-by-Step Solution

Answer

The polynomial can be factored as \(x(x - 3)(x + 3)(x^2 + 2)\).

Step 1: Factor out the Greatest Common Factor (GCF)

Look for the greatest common factor of the terms in the polynomial. In this case, each term contains an 'x', so we can factor out 'x' from the polynomial:\[x \left(x^4 - 7x^2 - 18\right)\]

Step 2: Substitute and Factor the Quadratic Polynomial

Take the quadratic polynomial \(x^4 - 7x^2 - 18\). Substitute \(y = x^2\) to transform it into a quadratic form \(y^2 - 7y - 18\). Now, factor this quadratic expression.The factors are: \(y^2 - 7y - 18 = (y - 9)(y + 2)\).

Step 3: Substitute Back the Original Variable

Substitute back \(y = x^2\) to get the expression back in terms of 'x'. Therefore:\((x^2 - 9)(x^2 + 2)\).

Step 4: Factor the Difference of Squares

Notice that \(x^2 - 9\) is a difference of squares. It can be factored further:\(x^2 - 9 = (x - 3)(x + 3)\).

Step 5: Write the Final Factored Form

Combine all the factors identified:\[x(x - 3)(x + 3)(x^2 + 2)\]This expression is now in the form of \(a_n(x-r_1)(x-r_2)\cdots(x-r_n)\) with \(a_n = 1\), \(r_1 = 0\), \(r_2 = 3\), and \(r_3 = -3\) with \(x^2 + 2\) irreducible over the reals.

Key Concepts

Greatest Common Factor

The concept of the Greatest Common Factor (GCF) is crucial when dealing with polynomials. It involves identifying the largest term that divides all terms of the polynomial without a remainder. By factoring out this term, we simplify the polynomial and make further factorization more manageable.
In the exercise, we begin by identifying 'x' as the GCF of the polynomial terms \(x^5, -7x^3,\) and \(-18x\). Each term contains at least one 'x', so we factor 'x' out.

• This reduces the polynomial to \(x(x^4 - 7x^2 - 18)\).
• By factoring out the GCF first, we simplify the polynomial into a form where additional techniques, like factoring quadratics, can be applied more efficiently.

Quadratic Polynomials

Quadratic polynomials are equations of the form \(ax^2 + bx + c = 0\). They play a pivotal role when factoring polynomials. In our exercise, once we have factored out the GCF, we encounter the expression \(x^4 - 7x^2 - 18\). While this is not initially in quadratic form, we can transform it into one by using a substitution technique.
By substituting \(y = x^2\), the expression becomes \(y^2 - 7y - 18\). This is now a quadratic equation that we can factor.

• The factors in this case are \((y - 9)(y + 2)\).
• We then substitute back \(y = x^2\), resulting in \((x^2 - 9)(x^2 + 2)\), bringing the factorization back to our original variable.

Difference of Squares

The difference of squares is a special factoring case. It applies when a polynomial can be expressed as \(a^2 - b^2\), which factors into \((a - b)(a + b)\). This technique is powerful because it simplifies expressions that look complex at first glance.
In the provided exercise, after substituting back to \((x^2 - 9)\), we notice that this term is a difference of squares. Here, \(x^2\) is \(a^2\) and \(9\) is \(b^2 = 3^2\). Thus, it can be factored as \((x - 3)(x + 3)\).

• This simplification is essential for expressing the polynomial in a factored form, revealing its roots and structure.
• Using the difference of squares aids in reducing and unraveling polynomial expressions efficiently.