Factor by any method. See Examples 1–7.

Question

q^{2} + 6 q + 9 - p^{2}

Accepted Answer

Hey everyone here we are asked to factor the following polynomial of Q squared plus 12 Q plus 36 minus four P squared. Now, before we start factoring this expression here, we want to begin by grouping together our first three terms. So we have the quantity of Q squared plus Q plus 36. And then we have this minus four P squared. And so now we just want to focus on our group terms for now. And doing this, we can see that we can apply the identity where we have X squared plus two, xy plus Y squared, which is equal to the quantity of X plus y squared. So now we just want to rewrite our grouped terms here so that it better matches our formula. And so we have Q squared plus 12 Q plus 36. And rewriting. So it matches the formula we have Q squared plus two times Q times six plus six squared. So now we can I compare this expression this rewritten expression with our formula to identify the values for X and Y. So first looking for our value for X. Well we see that from our formula that in the middle term, the next to the coefficient of two, we see we have our X term. So in our expression we see that X is going to be Q. And then again looking at our formula, we see that Y is right next to X in the middle term. So our Y. From our expression is going to be six and now with these values for X and Y. Clearly identified, we can go ahead and just substitute them in to the right hand side of our formula. So we see that our expression of Q squared plus 12 Q plus 36 takes the factored form according to our expression of the quantity of Q plus six squared. And so now we can take this factored quantity that we found and substitute it back into our original expression from earlier where we had the quantity of Q squared plus 12 Q plus 36 minus four P squared. And now substituting in our factored expression that we found for our group terms. We can rewrite our expression as the quantity of Q plus six squared minus four P squared. And now looking at our rewritten expression here, we see that we have the difference of two squares. So if we recall the difference of squares formula which is x squared minus y squared is equal to the quantity of X plus Y times the quantity of x minus y. Now we can rewrite our expression here so that it better matches our difference of squares formula. And doing this, we have the quantity of Q plus six squared minus the quantity of two P squared. And so now again we have an expression that perfectly matches the formula. So we can compare the expression with the formula to find our values for X and Y. Well from our formula, we see that our first term is going to be x squared from our expression. So our X squared in this case is the quantity of Q plus six squared. And this tells us that our value for X is just going to be Q plus six. And now again looking at our formula, we see that y squared is our second term from our expression. So y squared is going to be the quantity of two P squared, which tells us that y is just going to be two P. And now, once again, since we have identified the values for X and Y, we can go ahead and substitute them in to the right hand side of our formula. And so we see that our expression from earlier where we had the quantity of Q plus six squared minus the quantity of two P squared. We see that the factored form according to our formula becomes the quantity of Q plus six plus two p times the quantity of Q plus six minus two P. And now taking a look at our final factored expression here we see that there aren't any other terms to factor out here. So this is our final factored form of our original expression which leaves us with answer choice. C Thanks for watching. I hope you found this video helpful and I'll see you again next time

Factor by any method. See Examples 1–7. $q^{2} + 6 q + 9 - p^{2}$

Key Concepts

Factoring Quadratic Trinomials

Difference of Squares

Combining Factoring Methods

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