Problem 43 - Elementary Algebra

Question

Use the \(a c\) method to factor. Check the factoring. Identify any prime polynomials. $$ x^{2}+20 x+99 $$

Step-by-Step Solution

Answer

The factored form is \((x+11)(x+9)\). It is not a prime polynomial.

Step 1 - Identify coefficients

Identify the coefficients from the quadratic equation. For the equation \(x^{2}+20x+99\), the coefficients are as follows:\(a=1\)\(b=20\)\(c=99\)

Step 2 - Calculate the product of a and c

Next, calculate the product of \(a\) and \(c\):\(a*c=1*99=99\)

Step 3 - Find two numbers that multiply to a*c and add to b

We need to find two numbers that multiply to 99 (\(a*c\)) and add to 20 (\(b\)). The correct pair of numbers is 11 and 9, since:\[11 \times 9 = 99\]\[11 + 9 = 20\]

Step 4 - Rewrite the middle term

Rewrite the middle term using the pair of numbers found:\(x^{2} + 20x + 99\) becomes \(x^{2} + 11x + 9x + 99\)

Step 5 - Factor by grouping

Group the terms and factor each group:\(x^{2} + 11x + 9x + 99\)Group as:\((x^{2} + 11x) + (9x + 99)\)Factor each group:\(x(x+11) + 9(x+11)\)

Step 6 - Factor out the common binomial

Factor out the common binomial factor:\((x+11)(x+9)\)

Step 7 - Check the factoring

Multiply the factors to check:\((x+11)(x+9) = x^{2} + 9x + 11x + 99\)\(= x^{2} + 20x + 99\)The original equation is obtained, so the factoring is correct.

Step 8 - Identify if prime polynomial

Since the polynomial \(x^{2} + 20x + 99\) is factored successfully, it is not a prime polynomial.

Key Concepts

AC method

The AC method is a technique used to factor quadratic equations of the form \(ax^2 + bx + c\). It is particularly useful when the leading coefficient \(a\) is not equal to 1. The method involves several steps:

First, identify the coefficients of the quadratic equation. For example, in the equation \(x^2+20x+99\), the coefficients are \(a = 1\), \(b = 20\), and \(c = 99\).
Next, calculate the product of \(a\) and \(c\). This helps you find two numbers whose product equals \(ac\) and whose sum equals \(b\). In our example, \(ac = 1*99 = 99\).
The goal is to find two numbers that multiply to 99 and add up to 20. These numbers will help decompose the middle term, making it easier to factor by grouping. In this case, the numbers are 11 and 9, because \(11 \times 9 = 99\) and \(11 + 9 = 20\).

Rewrite the original quadratic equation by breaking the middle term using the identified numbers.\(x^2+20x+99\) becomes \(x^2 + 11x + 9x + 99\). Finally, factor by grouping, as we will explore in the related sections about quadratic equations and factor by grouping. Once you separate into groups, factor each group and then factor out the common binomial.

Quadratic Equation

Quadratic equations take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) represents an unknown variable. These equations have a degree of 2, which means the highest exponent of the variable is 2.
In general, the graph of a quadratic equation is a parabola, which can either open upwards or downwards depending on the sign of the leading coefficient \(a\).

To solve a quadratic equation, one common method is factoring. Factoring involves breaking down the equation into two binomials, which when multiplied together give you the original equation. For example, the equation \(x^2+20x+99\) can be factored into \((x+11)(x+9)\). To confirm the factoring is correct, we multiply the binomials:
\( (x+11)(x+9) \)
= \(x^2 + 9x + 11x + 99\)
= \(x^2 + 20x + 99\)
Since the product matches the original equation, the factoring is accurate.

Factoring works well for many quadratic equations, but not all can be factored easily. In such cases, you may need to use other methods, such as completing the square or the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Factor by Grouping

Factor by grouping is a useful technique for factoring polynomials, especially when dealing with quadratic equations. This method involves organizing the terms into smaller groups and factoring out common factors from each group.

Let's revisit our previous example of \(x^2 + 11x + 9x + 99\). We can group the terms as follows:
\((x^2 + 11x) + (9x + 99)\)
Here, you notice that each group has a common factor. Factor out these common factors from each group:
\(x(x + 11) + 9(x + 11)\).
Now, you can see that \(x+11\) is a common binomial factor. Factoring this out gives us:
\((x+11)(x+9)\).

To check your work, multiply the factors to see if you retrieve the original equation.
\((x+11)(x+9) = x^2 + 9x + 11x + 99 = x^2 + 20x + 99\).
This confirms that the factoring is correct. By learning to factor by grouping, you will be better equipped to simplify and solve quadratic equations.