In Exercises 49–64, factor any perfect square trinomials, or stat...

Question

In Exercises 49–64, factor any perfect square trinomials, or state that the polynomial is prime.

x⁴ − 4x² + 4

Accepted Answer

Welcome back. I am so glad you're here. We're asked to completely factorize the following perfect square trinomial. If possible, otherwise declare if the expression is a prime polynomial, our expression is s raised to the fourth power minus eight S squared plus 16. Our answer choices are answer choice A S plus two squared multiplied by S minus two squared. Answer choice B S plus four squared multiplied by S minus four squared. Answer choice C S squared minus four squared and answer choice D this polynomial is prime. All right, let's see if we do have a perfect square trinomial here and then could in fact factor it. So we know from previous lessons that a perfect square trinomial is expressed as A squared plus two A B plus B squared. So we're going to set each part of our given trinomial equal to each part of the perfect square trinomial and see if they match up. So first we'll set S rays to the fourth power equal to A squared. And then we're going to solve for A and substitute that in this two A B in the middle. So solving for A squared, we can take the square root of both sides and the square ret of S rays to the fourth power is S squared, that's equal to, on the right hand side, the square root of A squared is A. So we have solved for A, now we're going to solve for B. So this time we will set B squared equal to positive 16. So we have 16 equals B squared. Solving for B again will take the square rot of both sides of the equation and the square rot of 16 is plus or minus four. And that's equal to the square of B squared, which is B on the right hand side. So now we are going to set our negative eight S squared equal to this two A B in the middle here. And we'll see if they match. But we've actually got two BS because we have the plus or minus four. So I'm going to rate two of these. So we also have negative eight S squared equals another two A B. Now let's substitute in our A's and B's. So A is S squared. So I'll bring this down negative eight S squared equals two, multiplied not by A, but now by S squared and by B, we'll go with positive four for this first one. Now setting up the second equation negative eight S squared equals two multiplied by not A but S squared. And then instead of B, we'll go with the negative four. Now seeing if these two sides are the same. Let's work on the equation on the left, we have negative eight S squared equals we multiply our two and our four. And that's going to give us a positive eight and then bring down the S squared, those two do not match. So our B does not equal a positive four. Now, let's see if the equation on the right matches, we can bring down negative eight squared equals and then on the right, we'll multiply two and negative four and that equals a negative eight and then bring down our S squared and this one does match. So A equals S squared and B equals negative four. We do have a perfect square trinomial and here is how we factor it. We know that our perfect square trinomial is equal to open parentheses. A plus B close parentheses squared. So we just need to plug in our A S and BS. I'll bring this down here. We have open parentheses. A plus B closed parentheses squared. Now plugging in for A, we'll put in S squared and plugging in for B, we have a negative four, enclose those in parentheses and square it. And this is looking good this matches with answer choice C. However, we notice that we have a difference of squares. So there's more we can do here. All right, we recall from previous lessons that a difference of squares is written as A squared minus B squared. And we have that once we factor it, we will have open parentheses. A plus B closed parentheses, multiplied by open parentheses. A minus B closed parentheses. And I just want to note here that when we say we have open parentheses, S squared minus four, closed parentheses, squared, we're saying we've got two of these. So in open parentheses, S squared minus four closed parentheses multiplied by open parentheses as squared minus four, close parentheses. So each of these two are a difference of squares. Let's identify our A S and our BS. So if we set A squared equal to the first one S squared, and then we solve for our A, we take the square root of both sides and we'll get that A is equal to S. Now we want to solve for B squared, we set B squared equal to the second term in the parentheses, which is four and we take the square root of both sides and we get B equals two. And so we will have now plugging in A will be as so we have open parentheses, not A but S then plus now B plugging in for B will be two. So S plus two and one set of parentheses multiplied by another set of parentheses with, instead of A will be S then minus and instead of B will be two closed parentheses. But we have two of those because we have two sets of parentheses with S squared minus four. So this is multiplied by another S plus two, multiplied by S minus two. And now we can condense this a little bit. We've got two S plus twos and we have two S minus two factors. So we can set this equal to open parentheses. S plus two closed parentheses squared, multiplied by open parentheses. S minus two closed parentheses squared. And that is this one fully factored. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

In Exercises 49–64, factor any perfect square trinomials, or state that the polynomial is prime. x⁴ − 4x² + 4

주요 개념

Perfect Square Trinomials

Factoring Polynomials

Prime Polynomials

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