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Chapter 10: Problem 43

Factor by using trial factors. $$15 y^{2}-50 y+35$$

Short Answer

Expert verified
The factored form of the expression \(15 y^{2}-50 y+35\) is \( (5y -7 )(3y -5)\).

Step by step solution

01

Identify the coefficients

From the given quadratic equation, the coefficients are \(a = 15\), \(b = -50\), and \(c = 35\). We need to find two numbers that multiply to \(a*c = 15*35 = 525\) and add up to \(b = -50\).
02

Find the factors

The two numbers that satisfy these conditions are -25 and -21 as \(-25*-21=525\) and \(-25+-21=-50\). Therefore, break up the middle term of the quadratic equation using these factors. The expression becomes \(15 y^{2} - 25y - 25y + 35\).
03

Group the terms

We then group the terms for easier factoring: \( (15 y^{2} -25y) - (25y -35)\).
04

Factor by grouping

Factor each group separately. The expression becomes \(5y (3y -5) -7 (5y -7)\).
05

Factor out the common binomial

Finally, factor out the common binomial term \((3y -5)\) to obtain \( (5y -7 )(3y -5)\).

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Key Concepts

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

Quadratic Expressions
A quadratic expression is any mathematical expression of the form \( ax^2 + bx + c \). This is known as a quadratic polynomial because the highest exponent of the variable \( x \) is 2. Quadratic expressions are a common type of polynomial and initially appear in algebra.

When working with these expressions, it's crucial to recognize the structure:
  • \( a \) is the coefficient of \( x^2 \)
  • \( b \) is the coefficient of \( x \)
  • \( c \) is the constant term
Understanding this structure helps in solving quadratic equations through methods like factoring, completing the square, and using the quadratic formula.
Coefficient Identification
Coefficient identification involves finding the values of \( a \), \( b \), and \( c \) in a quadratic expression \( ax^2 + bx + c \). This step is crucial, as these coefficients impact which factoring method is the best to use.

To identify the coefficients:
  • Look at the term with \( x^2 \): the number in front is \( a \).
  • Observe the term with \( x \): the number in front is \( b \).
  • The standalone number is \( c \).
For example, given the expression \( 15y^2 - 50y + 35 \), the coefficients are:\\( a = 15 \), \( b = -50 \), \( c = 35 \). Proper identification is the first step in trial factors or other factoring methods.
Factor by Grouping
Factor by grouping involves rearranging and pairing terms in a quadratic expression to make it easier to factor. This method relies heavily on algebraic manipulation and is useful when direct factoring isn't obvious.

Here's how to use this method:
  • Rewrite the quadratic expression by splitting the middle term using the factor pair found during examination.
  • Group the resulting terms into pairs.
  • Factor out the common factor from each pair.
By following these steps, the expression \( 15y^2 - 50y + 35 \) can be rewritten as \( 15y^2 - 25y - 25y + 35 \), grouped further, and then common factors can be easily identified and factored out.
Trial Factors
Trial factors is a method of finding factors of a quadratic expression by examining possible pairs of numbers that multiply to give \( a*c \) and add up to \( b \).

The steps are as follows:
  • Calculate \( a*c \) from the coefficients of the quadratic expression.
  • Identify all pairs of factors of the product \( a*c \) and find which pair sums to \( b \).
  • Use the found factor pair to split the middle term and proceed with factoring.
For example, with \( 15y^2 - 50y + 35 \), compute \( a*c = 15*35 = 525 \). The pair \(-25\) and \(-21\) multiplies to 525 and adds to \(-50\), allowing us to factor the expression successfully.

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Most popular questions from this chapter

Given that \(x+2\) is a factor of \(x^{3}-2 x^{2}-5 x+6,\) factor \(x^{3}-2 x^{2}-5 x+6\) completely.

Factor. $$4(y-1)^{2}-7(y-1)-2$$

$$\text { If }-2 x^{3}-6 x^{2}-4 x=a(x+1)(x+2), \text { find } a$$

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