In Exercises 83–92, factor by introducing an appropriate substitu...

Question

In Exercises 83–92, factor by introducing an appropriate substitution.

3(x−2)² − 5(x−2) − 2

Accepted Answer

Hello, today we're gonna be using the substitution method to factor the given trinomial. So what we are given is four times the quantity X minus three squared plus 49 times the quantity X minus three minus 39. Now, if we try to use the trial and error method to factor the trinomial, the way that it is, it's gonna become a very tedious and extensive process. But notice that we have two instances of the group X -3. If we want to use a substitution, we can let a variable like a equal to X -3. And that can allow us to replace these X -3 quantities with a. So if we apply the substitution to the current trinomial, what this gives us is four A squared plus 49 A minus 39. And now we can go ahead and factor this using the trial and error method. So the trial and error method asks us to write out the factors of the first term and the last term, The factors of the first term which are called the P factors are just all the possible factors of only positive numbers. So what are going to be the possible factors of four a. Well, that can be four A times A or it could be two a times two A. And for the factors of the last term, which I'm going to call the Q factors, these are gonna be the list of factors for the last term which can be a combination of positive or negative values. So what are two numbers that we can multiply together to give us -39? Well, we can use one and negative 39 or negative one and positive 39. We can also choose to use three and negative or negative three and positive 13. So now that we have these factors listed out, I'm gonna go ahead and create two pairs of parentheses in these pairs of parentheses. The first term is going to be the P factors and the second terms are going to be the Q factors which can be positive or negative. So now let's go ahead and pick a combination of P factories and Q factors to test out. Let's go ahead and try four a with a And let's go ahead and try negative three with positive 13. So if we plug this into our groups, what we get is four, a -3 times a plus 13. And in order to verify that this is the factored form of the trinomial, we're gonna multiply the first term with the very last term and the two middle two terms together. And if the sum of their products add up to give us this middle term A, then this is going to be the correct factorization for the trinomial. So four A times 13 is going to give us positive 52 a and negative three times A is going to give us negative three A And 52 a -3 a is going to give us positive 49 a which matches the middle term of the trinomial. So that means that four a - times a plus 13 is going to be the correct factorization of the given trinomial. But we have one more step that we need to do. Recall that in the beginning, we let a equal to X -3. So now what we need to do is we need to ress substitute this a back into the two factorized groups. And doing this is going to give us four times the quantity X minus three minus three times X minus three plus 13. Now, we just need to go ahead and simplify the groups. What we can do in the left group is distribute the four into X minus three and that is going to give us four X minus 12 minus three. And on the right hand side, negative three plus 13 is going to give us positive 10. So we have X plus 10. Now, one more simplification to do negative 12 minus three is going to give us negative 15. So what we are left with is four X minus 15 times the quantity X plus 10. And this is going to be the completely factorized form of the original trinomial. And with that being said, the answer to this problem is going to be c so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 83–92, factor by introducing an appropriate substitution. 3(x−2)² − 5(x−2) − 2

Key Concepts

Factoring Polynomials

Substitution Method

Quadratic Expressions

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