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Chapter 6: Problem 77

In Exercises \(75-82,\) factor completely. $$(a+b) x^{2}+(a+b) x-20(a+b)$$

Short Answer

Expert verified
The completely factored form of the given polynomial is \((a+b)(x + 5)(x - 4)\)

Step by step solution

01

Identify Common Factor

Inspect the polynomial for common factors in all terms. In this case, it can be observed that \((a+b)\) is a common factor among all terms. The common factor can be factored out, simplifying the equation to: \((a+b)(x^{2} + x - 20)\)
02

Factor the Quadratic

Next, the quadratic expression \(x^{2} + x - 20\) needs to be factored. The factors of the quadratic equation are determined by finding two numbers that multiply to -20 (the constant term) and add up to 1 (the coefficient of x). The numbers satisfying these criteria are 5 and -4. Thus, this quadratic equation can be factored as \(x^{2} + x - 20 = (x + 5)(x - 4)\)
03

Final Factoring

Combine the factored common factor from step 1 with the factors from the quadratic. This gives us the completely factored form of the original equation as: \((a+b)(x + 5)(x - 4)\)

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Key Concepts

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

Common Factor
Every polynomial may contain one or several common factors across its terms. Identifying these is a fundamental step in the process of factoring polynomials. A common factor is a term that evenly divides each term within the polynomial. For instance, consider the given polynomial (a+b)x^2 + (a+b)x - 20(a+b). Upon scrutiny, one notices that each term contains the binomial (a+b) which is deemed the common factor. This factor can be 'factored out', leading to a simplified expression that is easier to work with.

Factoring Quadratic Equations
After extracting the common factors, one often encounters quadratic equations鈥攚hich are represented by a standard form ax^2 + bx + c. There are various methods to factor these, like using the FOIL method, completing the square, or applying the quadratic formula. However, when the coefficients are suitable, locating two numbers that multiply to yield the constant term c, and sum up to the coefficient b, allows for quick factoring into two binomials. In the given exercise, x^2 + x - 20 was broken down into (x + 5)(x - 4) by finding the numbers 5 and -4 which satisfy the multiplication and addition requirements.

Polynomial Expressions
A polynomial expression is a mathematical phrase which can have constants, variables and the exponents of those variables, where exponents must be whole numbers. In the context of the problem they range from monomials with one term, to binomials with two, trinomials with three, and so on. The exercise given features a trinomial鈥攖hree terms that include variable powers up to the second degree. Factoring polynomials is akin to breaking down numbers into their prime factors; it simplifies multiplication and allows for clearer understanding of the polynomial鈥檚 behavior.

Algebraic Techniques
Factoring polynomials is an essential algebraic technique that facilitates the simplification and solving of equations. These techniques are myriad and often intertwine, such as extracting common factors, foiling, expanding, or using special product rules and formulas. Mastering them is a bit like assembling a toolbox; the more tools you have and the better you know how to use them, the more types of equations you'll be able to tackle. In the given problem, identifying the common factor and factoring the quadratic are two such techniques that lead us to the solution. As demonstrated, using such techniques step-by-step makes complex problems more manageable.

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Most popular questions from this chapter

Solve: \(5 x+28=6-6 x\). (Section \(2.2,\) Example 7 )

Graph using intercepts: \(5 x-2 y=10\) (Section \(3.2,\) Example 4 )

Factor the numerator and the denominator. Then simplify by dividing out the common factor in the numerator and the denominator. \(\frac{x^{2}+6 x+5}{x^{2}-25}\)

Use the \(x\)-intercepts for the graph in \(a[-10,10,1]\) by \([-13,10,1]\) viewing rectangle to solve the quadratic equation. Check by substitution. Use the graph of \(y=x^{2}+x-6\) to solve \(x^{2}+x-6=0\).

In Exercises \(137-141,\) factor completely. $$4 x^{4}-9 x^{2}+5$$
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