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Chapter 5: Problem 167

Factor each polynomial using the trial-and-error method. $$ 5 y^{2}-21 y+4 $$

Short Answer

Expert verified
(5y - 1)(y - 4)

Step by step solution

01

- Identify the general form

The polynomial is in the quadratic form: \[a y^{2}+b y + c\]. Here, \(a = 5\), \(b = -21\), and \(c = 4\).
02

- Propose factor pairs of the quadratic term and constant term

List the factor pairs of the quadratic coefficient (5) and the constant term (4): Factor pairs of 5: (1,5) and (5,1) Factor pairs of 4: (1,4), (2,2), and (4,1).
03

- Trial and error with factor pairs

We test combinations from these pairs that sum to the middle term coefficient (-21). The form will be: \[(my + n)(py + q)\]. Testing simple pairs first, say \((5y + 1)(y + 4)\): We calculate the middle term as follows: \[5y \times 4 + 1 \times y = 20y + y = 21y \]. Adjust the signs: Testing \((5y - 1)(y - 4)\): We calculate the middle term as follows: \[5y \times -4 + (-1)y = -20y - y = -21y \]This product yields our middle term, confirming the factorization.
04

- Verify the factorization by expanding

Expand the factorized form to check: \[(5y - 1)(y - 4) = 5y \times y - 5y \times 4 - 1 \times y + (-1) \times (-4)\]\[= 5y^{2} - 20y - y + 4\]\[= 5y^{2} - 21y + 4\]. This matches the original polynomial, verifying that the factorization is correct.

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Key Concepts

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

quadratic formula
The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). The formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a\), \(b\), and \(c\) are coefficients from the original equation.
Here鈥檚 how it works:
  • Identify the coefficients \(a\), \(b\), and \(c\) in your quadratic equation.
  • Substitute these values into the quadratic formula.
  • Calculate the discriminant \(\Delta = b^2 - 4ac\).
  • Evaluate the two possible solutions using the \( \pm \) in the formula.
This formula helps find the roots of quadratic equations directly. Unlike other methods like factoring, the quadratic formula always works, even if the equation cannot be factored neatly.
trial-and-error method
The trial-and-error method is a straightforward way to factor quadratic polynomials when other methods may be unclear.
Let鈥檚 break it down:
  • First, identify the structure of your quadratic: \(ay^2 + by + c\).
  • Next, list the factor pairs of the quadratic coefficient (\(a\)) and the constant term (\(c\)). For example, for the quadratic \(5y^2 - 21y + 4\), the coefficient pairs are (1, 5) and (5, 1) and the constant pairs are (1, 4), (2, 2), and (4, 1).
  • Test various combinations of these factor pairs in the form \((my + n)(py + q)\).
  • Adjust signs and terms to match the middle term. This might take several tries.
While this method can be a bit time-consuming, it鈥檚 very practical for understanding polynomial factorization visually and step-by-step.
polynomial factorization
Polynomial factorization involves breaking down a polynomial into simpler polynomials that, when multiplied together, will give you the original polynomial.
Here are some key points:
  • Factoring helps find the roots of the polynomial, which are the values of \(y\) that make the polynomial equal to zero.
  • It simplifies complex algebraic expressions and makes them more manageable.
For our example \(5y^2 - 21y + 4\), the factorization was found to be \((5y - 1)(y - 4)\). This was verified by expanding the factors and checking if they multiply back to the original polynomial.
Polynomial factorization is a fundamental skill in algebra that applies to solving equations, simplifying expressions, and even calculus.

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Most popular questions from this chapter

Finding numbers. If the square of a number decreased by the number is \(12,\) then what is the number?

Factor each polynomial by grouping. $$ b m-4 b-5 m+20 $$

Consecutive integers. The sum of the squares of two consecutive integers is \(25 .\) Find the integers.

Factor each polynomial using the trial-and-error method. $$ a^{2}-8 a-33 $$

Solve each equation. $$(2 x-1)\left(x^{2}-9\right)=0$$
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