Problem 57 - Introductory Algebra for College Students

Question

Factor completely. $$y^{4}+2 y^{3}-80 y^{2}$$

Step-by-Step Solution

Answer

The factorized form of the given equation is \(y^{2}(y+10)(y-8)\)

Step 1: Identify common factor

From the given expression \(y^{4}+2 y^{3}-80 y^{2}\), it can be seen that \(y^{2}\) is a common factor in all terms.

Step 2: Factor out the common factor

The next step is to factor out the common factor from all terms. This gives us: \(y^{2}(y^{2}+2y-80)\)

Step 3: Factor the quadratic polynomial

The quadratic polynomial inside the brackets can be further factored using common techniques of factoring a quadratic polynomial of form \(ax^{2} + bx + c\). Two numbers are being looked for that multiply to -80 (product ac) and add to 2(sum b). These numbers are 10 and -8. So, the expression becomes: \(y^{2}[(y+10)(y-8)]\)

Key Concepts

Common Factor

Factoring expressions often starts with identifying a 'common factor.' This is the largest factor that is shared by each term in the expression. Identifying the common factor can significantly simplify the expression, making further factoring more straightforward.
It's like finding a common thread that links each term together. In the expression given, \(y^{4}+2y^{3}-80y^{2}\), each term includes \(y^{2}\) as a factor. By pulling out \(y^{2}\), you simplify the expression to a more manageable form: \(y^{2}(y^{2}+2y-80)\).
By factoring out a common factor, you essentially divide each term in the polynomial by \(y^{2}\). This marks the initial step towards breaking down complex algebraic expressions into simpler components.

Quadratic Polynomial

The next step involves working with a quadratic polynomial, a common expression of the form \(ax^2 + bx + c\). In our example, after factoring out the common factor, you are left with \(y^{2}+2y-80\). This is a typical quadratic polynomial and requires specific methods to factor further.
Quadratic polynomials are prevalent in high school mathematics because they elegantly represent situations involving areas, projectile motion, and many real-world phenomena. Factoring a quadratic helps solve equations of this type and turns complex problems into simpler multiplier ones.
When handling such polynomials, recognizing its structure allows you to employ various factoring techniques to break it into linear factors or smaller quadratics when possible.

Factoring Techniques

Once a quadratic polynomial is isolated, various techniques can be employed to factor it. In the example, you aim to rewrite \(y^{2}+2y-80\) as a product of simpler binomials. Here are some techniques to consider:

• Factoring by Inspection: Look for two numbers that, when multiplied, give the constant term \(c\) (-80 in this case) and when added, give the linear coefficient \(b\) (2 here).
• Trial and Error: Guess and check possible factors based on integer options for a and b.
• Using the Quadratic Formula: In more systematic approaches where inspection is difficult, this formula provides solutions for roots or potential factors.

In the current problem, the numbers 10 and -8 fit both requirements: they multiply to -80 and add to 2, leading to factors \((y+10)(y-8)\). Combining these steps, the fully factored form of the original polynomial becomes \(y^{2}(y+10)(y-8)\). Utilizing these techniques successfully breaks down what initially appears to be a complex scenario into simple multiplication, simplifying further calculations or solutions.