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Chapter 5: Problem 48

Factor. $$ 10 y^{2}+7 y-12 $$

Short Answer

Expert verified
(2y + 3)(5y - 4)

Step by step solution

01

- Identify coefficients

The trinomial is in the form of \(a y^2 + b y + c\). Here, \(a = 10\), \(b = 7\), and \(c = -12\).
02

- Find the product of \(a\) and \(c\)

Calculate the product of the first and third coefficients: \(a\cdot c = 10 \cdot (-12) = -120\).
03

- Find two numbers that multiply to \(ac\) and add to \(b\)

We need two numbers that multiply to \(-120\) and add to \(7\). These numbers are \(15\) and \(-8\) because \(15 \times (-8) = -120\) and \(15 + (-8) = 7\).
04

- Split the middle term

Rewrite the middle term using the two numbers found: \[10y^2 + 15y - 8y - 12\].
05

- Factor by grouping

Group the terms into pairs and factor out the common factors from each pair: \( (10y^2 + 15y) + (-8y - 12) \). This gives us \( 5y(2y + 3) - 4(2y + 3) \).
06

- Factor out the common binomial

Factor out the common binomial \((2y + 3)\): \[ (2y + 3)(5y - 4) \].

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Key Concepts

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

coefficients
In the given trinomial, the coefficients play a key role in the factoring process. Coefficients are the numerical factors in terms of a polynomial. For our trinomial, which is in the form of \(a y^2 + b y + c\), the coefficients are:
  • \(a = 10\): This is the coefficient of \(y^2\).
  • \(b = 7\): This is the coefficient of \(y\).
  • \(c = -12\): This is the constant term.
When factoring trinomials, identifying these coefficients correctly is crucial.
In our trinomial: \(10 y^{2}+7 y-12\), we start by clearly noting these values. We will use them throughout the factoring process.
product-sum factoring
One popular method to factor trinomials is called product-sum factoring. This involves finding two numbers that both multiply to give the product of \(a \text{ and } c\) and add up to \(b\).
In our example:
  • Calculate the product of the first and third coefficients: \(10 \times -12 = -120\).
  • Next, find two numbers that multiply to \(-120\) and add to \(7\).
  • After some thinking, we find that these numbers are \(15\) and \(-8\):
    • \(15 \times -8 = -120\)
    • \(15 + (-8) = 7\)
You split the middle term based on these numbers:
Rewrite \(7y\) as \(+15y - 8y\) leading to:\
\(10 y^{2}+15 y-8 y-12\).
factor by grouping
The next approach is called factor by grouping. This is where we group terms in pairs to factor them separately, ultimately leading to a common factor.
Let's break down our modified expression \(10 y^{2} + 15 y - 8 y - 12\):
  • First grouping: \((10 y^2 + 15 y)\)
  • Second grouping: \((-8 y - 12)\)
Next, factor out the greatest common factor (GCF) from each pair:
  • \(5 y(2 y + 3)\) from the first group.
  • \(-4(2 y + 3)\) from the second group.
We observe that \((2 y + 3)\) is a common factor in both grouped terms.
Lastly, factor out this common binomial factor to get:
\((2 y + 3)(5 y - 4)\).
This is the factored form of our original trinomial \(10 y^{2} + 7 y - 12\).
The factor by grouping method simplifies complex trinomials by breaking them into manageable steps, ensuring accuracy and clarity.

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