Problem 17 - College Algebra with Corequisite Support

Question

For the following exercises, factor the polynomial. \(10 h^{2}-9 h-9\)

Step-by-Step Solution

Answer

The factorization of the polynomial is \((2h - 3)(5h + 3)\).

Step 1: Identify the Coefficients

The polynomial we need to factor is given by \(10h^2 - 9h - 9\). The coefficients are: \(a = 10\), \(b = -9\), and \(c = -9\).

Step 2: Calculate the Product \(ac\)

Compute the product of \(a\) and \(c\): \[ ac = 10 imes (-9) = -90 \]

Step 3: Find Two Numbers That Multiply to \(ac\) and Add to \(b\)

We need two numbers that multiply to \(-90\) and add to \(-9\). These numbers are \(6\) and \(-15\) because: \[ 6 imes (-15) = -90 \]\[ 6 + (-15) = -9 \]

Step 4: Rewrite the Middle Term

Use the two numbers found to split the middle term \(-9h\):\[ 10h^2 + 6h - 15h - 9 \]

Step 5: Factor by Grouping

Group the terms and factor each group separately:\[ (10h^2 + 6h) + (-15h - 9) \]Factor out the greatest common factor (GCF) in each group:\[ 2h(5h + 3) - 3(5h + 3) \]

Step 6: Factor Out the Common Binomial

Factor the common binomial factor \((5h + 3)\):\[ (2h - 3)(5h + 3) \]

Step 7: Verify the Factorization

Expand \((2h - 3)(5h + 3)\) to ensure it equals the original polynomial:\[ 2h(5h + 3) - 3(5h + 3) = 10h^2 + 6h - 15h - 9 = 10h^2 - 9h - 9 \]The factorization is correct.

Key Concepts

Coefficients in Polynomials

In the world of polynomials, coefficients hold a very important role. They are the numbers in front of the variables in a polynomial expression. For instance, in the polynomial \(10h^2 - 9h - 9\), you'll notice numbers right before the variables. These are the coefficients:

• \(a = 10\): This is the coefficient of \(h^2\).
• \(b = -9\): This is the coefficient of \(h\).
• \(c = -9\): This is the constant term, which also acts as a coefficient.

By identifying these coefficients, you gain insight into the formation of the polynomial. Understanding these values is crucial because they are used extensively in solving or factoring polynomials. Coefficients help in finding the right combination of terms when simplifying or resolving polynomial equations.

Polynomial Factorization Steps

Factoring polynomials involves breaking them down into simpler expressions that can be multiplied to give the original polynomial. Here are the organized steps:

• Identify the coefficients: Recognize the coefficients from the polynomial expression, which allow you to manage your calculations.
• Calculate the product \(ac\): Multiply the first coefficient (\(a\)) and the constant term (\(c\)). This product will aid in finding the right pair of numbers.
• Find suitable numbers: Look for two numbers that multiply to \(ac\) and add to the middle coefficient \(b\). This finds the split needed for factoring.
• Rewrite the middle term: Use the two numbers found to divide the middle term into two separate terms. This prepares the expression for grouping.
• Factor by grouping: Split the polynomial into two groups and factor each separately; the common factors will align for the last step.
• Factor out the common binomial: Combine the resulting expressions into simpler binomials, achieving the factorization goal.

Following these steps meticulously is key in polynomial factorization. Each step allows you to reduce the polynomial into factors, simplifying the expression for further mathematical tasks.

Factor by Grouping

"Factor by grouping" is a powerful technique when dealing with polynomials of higher degrees, especially when polynomial terms initially do not appear to be easily factorable. The process is systematic:

• Split the polynomial into two parts: Begin by using the numbers found during the factorization step (here, \(6h\) and \(-15h\)) to break up the expression into two distinct sections, ensuring each part can be grouped.


• Group the terms: Arrange the terms in pairs: \((10h^2 + 6h)\) and \((-15h - 9)\), which makes it easier to spot common factors.


• Factor out the GCF: Within each group, identify and extract the greatest common factor (or GCF), such as \(2h\) from the first group and \(-3\) from the second. This brings out a common binomial factor across both groups.


• Combine using the binomial factor: Once factored, the binomial like \((5h + 3)\) emerges as a common factor, leading to the expression \((2h - 3)(5h + 3)\).

This approach essentially packages the polynomial into simpler parts, making further operations or verifications much easier to perform. Factor by grouping capitalizes on reorganizing elements to unveil simpler underlying structures.