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

Question

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

5x⁴ + 2x² − 3

Accepted Answer

Hello, today we're gonna be using substitution to factor the given trinomial. So what we are given is seven X to the fourth plus 17 X squared minus 12. Now, one thing to note about the first term is that we can go ahead and rewrite X to the fourth as X squared squared using the multi, using the exponential multiplication. We can go ahead and multiply the two exponents of two and two together to give us X to the fourth. So we can rewrite the current trinomial as seven X squared squared plus 17 X squared minus 12. And now notice that we have two instances of X squared, we can go ahead and apply a substitution to replace the X squared in this trinomial. So let's go ahead and say Y is equal to X squared. In this case, if we apply the substitution, we can rewrite the trinomial as seven Y squared plus 17 Y minus 12. And now we can go ahead and factor this using the trial and error method. So the trial and error method says that we need to write out the factors of the first term and the last term the factors of the first term, which are going to be the P factors are going to be all the positive factors of seven Y squared. So since seven is a prime number, there is only one possible fact set of factors for that term and that is going to be seven Y and Y. And now the term, the factors for the last term which is going to be negative 12, which I'm going to call the Q factors is going to be all the possible factors of the combinations of positive and negative values. So what are the possible factors for negative 12? Well, that can be negative one and 12 or positive one and negative 12. We can also try four, negative three or negative four and positive three and we can also try six and negative two or negative six and positive two. Now that we listed out all the factors, I'm gonna go ahead and write two pairs of parentheses and in these pairs of parentheses, the first term is going to be the numbers from the P factors. And the last term is going to be the numbers from the Q factors. And again, those can be positive or negative. So let's go ahead and pick a different combination of P and Q factors to test the factorization. We're going to use seven Y and Y and let's go ahead and test that with negative four and positive three. So plugging this into our parentheses. Is going to give us seven Y minus four times Y plus three. And now, in order to verify that this is the correct factorization, we're gonna multiply the first term and very last term together. And we're gonna multiply the two middle two terms together. And if the sum of the products adds up to give us 17 Y, which is the middle term here that verifies that this is the correct factorization. So seven Y times positive three is going to give us 21 Y and negative four times Y is going to give us negative four, Y 21 Y -4 Y is going to give us 17 y which matches the middle term of the trinomial. So that means seven Y minus four times Y plus three is going to be the correct factorization for the trinomial. But we have one more step to keep in mind, recall that in the beginning of the problem, we let Y equal to X squared. So we need to put this X back into the factors by res substituting the value for Y. And doing this is going to allow us to rewrite the factors as seven X squared minus four times X squared plus three, which is going to be the completely factored version of the given trinomial. And with that being said, the answer to this problem is going to be a. 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. 5x⁴ + 2x² − 3

Key Concepts

Factoring Polynomials

Substitution Method

Quadratic Form

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