Problem 33 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$-2 y^{2}+2 y+112$$

Step-by-Step Solution

Answer

The factored form of the polynomial \( -2y^2 + 2y + 112 \) is \( 2(-y + 7)(y - 8) \)

Step 1: Check and Remove Common Factors

Let's see if there is a common factor that can be divided out in \( -2y^2 + 2y + 112\). We can divide each term by 2 to simplify it to \(-y^2 + y + 56\)

Step 2: Rewrite Quadratic

The next step is to find two numbers that multiply to \(ac\) and add to \(b\), resulting in the quadratic \( ax^2 + bx + c \). In this case, our \(a\) is -1, our \(b\) is 1 and our \(c\) is 56. So we want numbers that multiply to -56 and add up to 1. The numbers 7 and -8 fit these criteria.

Step 3: Splitting the Middle Term

We can now rewrite our original polynomial, splitting our 'y' term into two: \(-y^2 + 7y -8y + 56 \)

Step 4: Factorization by Grouping

We now apply the method of factorization by grouping. Group the terms two at a time and factor out the common factor from each group : \( y(-y + 7) -8(-y + 7) \)

Step 5: Common Factor again

We can now factor out the common factor of \(-y + 7\) and assemble our final factored form: \( (-y + 7)(y - 8)\)

Key Concepts

Common Factor

Finding a common factor is crucial when factoring polynomials. It simplifies the expression and makes the next steps much easier. When dealing with a polynomial like \(-2y^2 + 2y + 112\), the first thing you want to do is examine each term to check if they share a common factor.

• In our example, you see that each term can be divided by 2. This is the common factor or the greatest common factor (GCF).
• When you factor out the GCF, you're left with an expression that's simpler and easier to work with: \(-y^2 + y + 56\).
• This step only requires basic division, and it's crucial because it reduces the size of the coefficients, making the next factoring steps simpler.

So, always start factoring by identifying the common factor. It's like finding the key to unlock a simpler problem!

Quadratic Equations

Quadratic equations are polynomials of degree 2, in the form ax^2 + bx + c. To factor them, especially when the quadratic is simple (a = 1), you look for two numbers that multiply to the product of a and c and add up to b.

In our example \(-y^2 + y + 56\), the coefficients are a = -1, b = 1, and c = 56. Here’s how we solve it:

• Multiply a and c: \(-1 \times 56 = -56\).
• Find two numbers that multiply to \(-56\) and add to 1 (our b). These numbers are -7 and 8.

The key is correctly identifying these numbers since they allow us to split the middle term. When working with quadratic equations, always consider these relationships between a, b, and c. This will help you factor the expression successfully.

Factorization by Grouping

Factorization by grouping is an effective method when you have a four-term polynomial. It involves grouping terms to make factoring easier. After rewriting the quadratic by splitting the middle term, as in the example:

\(-y^2 + 7y - 8y + 56\), you apply grouping.

• Take the first two terms: \(-y^2 + 7y\) and factor out the common factor, which is \(y\). This gives \(y(-y + 7)\).
• Now, take the last two terms: \(-8y + 56\). Factor out \(-8\), leading to \(-8(-y + 7)\).
• Notice the common binomial factor \(-y + 7\).

Once you find this common factor, factor it out of each group to achieve \((-y + 7)(y - 8)\).

This process might seem lengthy, but it's methodical. Once you've done it a few times, it becomes second nature. Always look for possible groupings that reveal a common factor, as this will help simplify many polynomials.