That is,the factored form of. Solution Because the third term is positive and the middle term is negative, we find two negative integers whose product is 6 and whose sum is We list the possibilities. When the first term of a trinomial is positive and the third term is negative,the signs in the factored form are opposite.
That is, the factored form of. Solution We must find two integers whose product is and whose sum is It is easier to factor a trinomial completely if any monimial factor common to each term of the trinomial is factored first. For example, we can factor. A monomial can then be factored from these binomial factors. However, first factoring the common factor 12 from the original expression yields.
Solution We find two positive factors whose product is 6y 2 and whose sum is 5y the coefficient of x. The two factors are 3y and 2y. When factoring, it is best to write the trinomial in descending powers of x. If the coefficient of the x 2 -term is negative, factor out a negative before proceeding. Solution We look for two integers whose product is 12 and whose sum is 5.
From the table in Example 1 on page , we see that there is no pair of factors whose product is 12 and whose sum is 5. In this case, the trinomial is not factorable. Skill at factoring is usually the result of extensive practice. If possible, do the factoring process mentally, writing your answer directly.
You can check the results of a factorization by multiplying the binomial factors and verifying that the product is equal to the given trinomial. In this section, we use the procedure developed in Section 4. As before, if we have a squared binomial, we first rewrite it as a product, then apply the FOIL method. As you may have seen in Section 4. When a monomial factor and two binomial factors are being multiplied, it is easiest to multiply the binomials first.
First, we consider a test to determine if a trinomial is factorable. Consider the following possibilities. The test to see if the trinomial is factorable can usually be done mentally. We illustrate by examples. We consider all pairs of factors whose product is 4. Since 4 is positive, only positive integers need to be considered.
The possibilities are 4, 1 and 2, 2. We consider all pairs of factors whose product is 3. Since the middle term is positive, consider positive pairs of factors only. The possibilities are 3, 1. We write all possible arrangements of the factors as shown. We select the arrangement in which the sum of products 2 and 3 yields a middle term of 8x.
The integers 4 and -3 have a product of and a sum of 1, so the trinomial is factorable. We now proceed. We consider all pairs of factors whose product is 6. Since 6 is positive, only positive integers need to be considered. Then possibilities are 6, 1 and 2, 3. We consider all pairs of factors whose product is The possibilities are 2, -1 and -2, 1. We write all possible arrange ments of the factors as shown.
We select the arrangement in which the sum of products 2 and 3 yields a middle term of x. With practice, you will be able to mentally check the combinations and will not need to write out all the possibilities. Paying attention to the signs in the trinomial is particularly helpful for mentally eliminating possible combinations. Solution Rewrite each trinomial in descending powers of x and then follow the solutions of Examples 3 and 4.
As we said in Section 4. We know that the trinomial is factorable because we found two numbers whose product is 12 and whose sum is 8. Those numbers are 2 and 6. This is the same result that we obtained before.
Some polynomials occur so frequently that it is helpful to recognize these special forms, which in tum enables us to directly write their factored form. Observe that. Often we must solve equations in which the variable occurs within parentheses.
We can solve these equations in the usual manner after we have simplified them by applying the distributive law to remove the parentheses. Parentheses are useful in representing products in which the variable is contained in one or more terms in any factor.
One integer is three more than another. If x represents the smaller integer, represent in terms of x. The larger integer. Five times the smaller integer. Five times the larger integer. Let us say we know the sum of two numbers is If we represent one number by x, then the second number must be 10 - x as suggested by the following table. In general, if we know the sum of two numbers is 5 and x represents one number, the other number must be S - x. The next example concerns the notion of consecutive integers that was consid- ered in Section 3.
The difference of the squares of two consecutive odd integers is The larger integer b. The square of the smaller integer c. The square of the larger integer. Sometimes, the mathematical models equations for word problems involve parentheses. We can use the approach outlined on page to obtain the equation. Then, we proceed to solve the equation by first writing equivalently the equation without parentheses.
One integer is five more than a second integer. Three times the smaller integer plus twice the larger equals Find the integers. Steps First, we write what we want to find the integers as word phrases. Then, we represent the integers in terms of a variable. In this section, we will examine several applications of word problems that lead to equations that involve parentheses.
Once again, we will follow the six steps out- lined on page when we solve the problems. The basic idea of problems involving coins or bills is that the value of a number of coins of the same denomination is equal to the product of the value of a single coin and the total number of coins. There are 16 more dimes than quarters. How many dimes and quarters are in the col- lection? Steps We first write what we want to find as word phrases. Then, we represent each phrase in terms of a variable. How much is invested at each rate?
Step 3 Next, we make a table showing the amount of money invested, the rates of interest, and the amounts of interest. Step 4 Now, we can write an equation relating the interest from each in- vestment and the total interest received.
The basic idea of solving mixture problems is that the amount or value of the substances being mixed must equal the amount or value of the final mixture. Steps We first write what we want to find as a word phrase.
Then, we represent the phrase in terms of a variable. Kilograms of 80c candy: x. Step 3 Next, we make a table showing the types of candy, the amount of each, and the total values of each. Step 3 Next, we make a table or drawing showing the percent of each solu- tion, the amount of each solution, and the amount of pure acid in each solution. Since I started with a fourth-degree polynomial, then I'll be left with a quadratic once I divide out these two given factors.
I can solve that quadratic by using the Quadratic Formula or some other method. I need to remember that I divided off a " 3 " when I solved the quadratic; it is still part of the polynomial, and needs to be included as a factor.
Then the fully-factored form is:. Top Return to Index. Stapel, Elizabeth. Accessed [Date] [Month] The "Homework Guidelines". Study Skills Survey. Tutoring from Purplemath Find a local math tutor. Cite this article as:. Contact Us.
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