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Chapter 5: Problem 118

Apply the special factoring rules of this section to factor each polynomial. \(y^{2}-0.36\)

Short Answer

Expert verified
\(y^{2} - 0.36 = (y - 0.6)(y + 0.6)\)

Step by step solution

01

- Recognize the Form

Identify the form of the polynomial. Notice that the polynomial is a difference of squares: \(y^{2} - 0.36\).
02

- Express Terms as Squares

Write each term as a square. \(y^{2}\) is already a square, and 0.36 can be written as \((0.6)^2\) because 0.6 squared is 0.36.
03

- Apply the Difference of Squares Formula

Use the difference of squares formula which states that \(a^2 - b^2 = (a - b)(a + b)\). Here, \(a = y\) and \(b = 0.6\), so the equation becomes: \(y^{2} - (0.6)^{2} = (y - 0.6)(y + 0.6)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

difference of squares
The 'difference of squares' is a special concept in algebra. When you see a polynomial like the one in the exercise, you can check if it forms this special pattern.
The difference of squares follows the rule: \[a^2 - b^2 = (a - b)(a + b)\].
It means you're subtracting one perfect square from another perfect square.
  • Examples of perfect squares: \( x^2, 4, 25, 0.36 \)
  • Example conforming to difference of squares: \( y^2 - 0.36 \)
In the provided exercise, the polynomial \( y^2 - 0.36 \) fits the difference of squares structure.
algebra
In algebra, we often deal with polynomials鈥攅xpressions involving variables and coefficients. One powerful algebraic tool is factoring.
Factoring means breaking down a complex expression into simpler ones. Remember:
  • Expressions and equations can often be simplified or solved through factoring.
  • Understanding special forms like the 'difference of squares' helps in quick factoring.
For the given exercise, recognizing \( y^2 - 0.36 \) as a 'difference of squares' simplifies the process. Now, let's see how polynomial factoring fits into this.
polynomial factoring
Polynomial factoring breaks down a polynomial into simpler factors that, when multiplied together, give back the original polynomial.
The special 'difference of squares' helps to factor polynomials quickly. Steps in the example:\[ y^2 - 0.36 \]:
  • First, identify it as a difference of squares: \( y^2 - (0.6)^2 \).
  • Then, apply the difference of squares formula: \( a^2 - b^2 = (a - b)(a + b) \).
  • For \( y = a \) and \( 0.6 = b \), substitute and get: \( (y - 0.6)(y + 0.6) \).
Resulting factors: \( (y - 0.6)(y + 0.6) \).
Algebra becomes powerful when you can identify these patterns and use these methods.

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Most popular questions from this chapter

If an object is projected upward with an initial velocity of \(128 \mathrm{ft}\) per sec, its height \(h\) in feet after t seconds is given by the quadratic equation $$ h=-16 t^{2}+128 t $$ Find the height of the object after each time listed. \(4 \mathrm{sec}\)

Factor each trinomial. \(36 x^{4}+59 x^{2} y+24 y^{2}\)

Apply the special factoring rules of this section to factor each polynomial. \(36 m^{2}-\frac{16}{25}\)

Factor each polynomial completely. \((a-b)^{3}-(a+b)^{3}\)

Factor each trinomial. \(-2 a^{2}-5 a b-2 b^{2}\)
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