factor completely. remember to look first for a common factor. if a polynomial is prime state this. x^2+2xy+y^2-9
Try factorizing the first three terms using the identity:
(a+b)² = a² + 2ab + b²
The resulting difference of two squares can be factorized using the identity:
(a²-b²) = a² - b²
Post your answer for a check if you are not sure.
To factor the given polynomial x^2 + 2xy + y^2 - 9, we will first check if there is a common factor. In this case, there is no common factor among all the terms.
Next, we will attempt to factor the polynomial as a perfect square trinomial. A perfect square trinomial is an expression of the form (a + b)^2 or (a - b)^2, where a and b are terms or expressions.
Looking at the given polynomial, we can observe that it resembles the form (a + b)^2. In particular, we notice that the first two terms (x^2 and 2xy) have a common variable 'x', and the last two terms (2xy and y^2) have a common variable 'y'.
Applying the perfect square trinomial formula (a + b)^2 = a^2 + 2ab + b^2, we can rewrite the polynomial as:
x^2 + 2xy + y^2 - 9
= (x + y)^2 - 9
Now we have a difference of squares. Recall that the difference of squares formula is a^2 - b^2 = (a + b)(a - b). In this case, a = (x + y) and b = 3 (since 9 = 3^2). Applying the formula, we can factor the polynomial as:
= [(x + y) + 3][(x + y) - 3]
Thus, the completely factored form of the polynomial x^2 + 2xy + y^2 - 9 is (x + y + 3)(x + y - 3).