Higher-Order Polynomials

The first step when factoring any polynomial should be to check if there's a common factor in all terms. If there is, factor it out first.

For example, consider the polynomial:

$P\left(x\right)=4x^4+12x^3-40x^2$

Looking at each term:

• First term: $4x^4$
• Second term: $12x^3$
• Third term: $-40x^2$

We can see that $4x^2$ is a common factor:

• $4x^4=4x^2·x^2$
• $12x^3=4x^2·3x$
• $-40x^2=4x^2·\left(-10\right)$

So we can factor the polynomial as:

$P\left(x\right)=4x^2(x^2+3x-10)$

Notice that the degree of the polynomial is still 4 (the highest power of x), but we've simplified it by factoring out the common term.

We could factor this further by factoring the quadratic expression inside the parentheses:

$P\left(x\right)=4x^2(x+5)(x-2)$

After factoring, we can now easily identify that this polynomial has roots at x = 0 (with multiplicity 2), x = -5, and x = 2.

Sometimes you might encounter polynomials where you can factor out a higher-degree term. For example:

$Q\left(x\right)=x^5-4x^3$

Here, we can factor out $x^3$:

$Q\left(x\right)=x^3(x^2-4)=x^3(x+2)(x-2)$

This tells us that Q(x) has roots at x = 0 (with multiplicity 3), x = -2, and x = 2.

The SAT may present polynomials in either factored form or standard form. If they're in standard form, look for common factors first, then try to factor any remaining quadratic expressions using the techniques you've already learned.

Sometimes, after factoring out common terms, you might be left with a quadratic expression that cannot be easily factored. In such cases, you can use the quadratic formula to find the roots.

Consider this example:

$f\left(x\right)=4x^3+16x^2+8x$

First, let's check if we can factor out a common term:

Looking at each term:

• First term: $4x^3$
• Second term: $16x^2$
• Third term: $8x$

We can see that $4x$ is a common factor:

$f\left(x\right)=4x(x^2+4x+2)$

Now we have a cubic polynomial factored as a common term multiplied by a quadratic expression. We immediately see that x = 0 is a root of the original polynomial (from the factor 4x).

The quadratic expression $x^2+4x+2$ cannot be factored using rational coefficients. Let's verify this by checking if it has rational roots.

Using the quadratic formula for $x^2+4x+2=0$:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-4\pm \sqrt{16-4·1·2}}{2·1}=\frac{-4\pm \sqrt{8}}{2}$

$x=\frac{-4\pm 2\sqrt{2}}{2}=-2\pm \sqrt{2}$

So the roots of the quadratic factor are $x=-2+\sqrt{2}$ and $x=-2-\sqrt{2}$, which are irrational numbers (approximately -0.586 and -3.414).

Therefore, the polynomial f(x) has three real roots:

• $x=0$ (from the factor 4x)
• $x=-2+\sqrt{2}$ (approximately -0.586)
• $x=-2-\sqrt{2}$ (approximately -3.414)

This example demonstrates how a higher-order polynomial can be partially factored by identifying common terms, and then the quadratic formula can be used to find the remaining roots if the remaining quadratic cannot be factored easily.

If a polynomial is written in factored form, you can easily identify its roots.

Factored form just means that the whole function is written as a product of two or more expressions.

Due to the Zero Product Property, if any one of these expressions equals zero, the whole function equals zero.

Thus, any value of x that makes any one of these expressions equal to zero is a root of the function.

For example, if we have the polynomial:

$f\left(x\right)=3x{(x-2)}^2(x+1)(x-3)$

The roots of this polynomial are x = 0, x = 2, x = -1, and x = 3, because each of these values makes one of the factors equal to zero.

The behavior of a polynomial graph at each root (x-intercept) depends on whether the multiplicity of the root is even or odd.

The multiplicity of a root is the power that the corresponding factor is raised to.

• Odd Multiplicity
  • If a factor has an odd power, the graph passes through the x-axis at that root. The graph crosses from one side of the x-axis to the other.
  • For example, in the function $f\left(x\right)={(x-2)}^2{(x+1)}^3(x-4)$:
    • We have odd roots at:
      • x = -1 with multiplicity 3
      • x = 4 with multiplicity 1
    • Thus, at -1 and 4, the graph crosses the x-axis.
    • We an even root at x = 2 with multiplicity 2 (we'll talk about even roots next)
• Even Multiplicity
  • If a factor has an even power, the graph touches the x-axis at that root but doesn't cross it. The graph "bounces" off the x-axis and stays on the same side.
  • For example, in the function $f\left(x\right)={(x-2)}^2(x+1){(x-4)}^4$:
    • We have even roots at:
      • x = 2 with multiplicity 2
      • x = 4 with multiplicity 4
    • Thus, at 2 and 4, the graph touches the x-axis and turns around, so that it doesn't cross the x-axis.
    • We have an odd root at x = -1, so the graph crosses the x-axis here.

Let's look at an example to illustrate this:

Consider the polynomial: $f\left(x\right)=(x+1){(2x-1)}^2{(x-2)}^3$

This polynomial has three distinct roots:

• x = -1 (multiplicity 1, odd)
• x = 0.5 (multiplicity 2, even)
• x = 2 (multiplicity 3, odd)

The graph of this polynomial would look something like this:

Notice how the graph behaves differently at each root:

• At x = -1, the graph crosses through the x-axis (odd multiplicity)
• At x = 0.5, the graph touches the x-axis but doesn't cross it (even multiplicity)
• At x = 2, the graph crosses through the x-axis (odd multiplicity)