Solve Notice that if you take ½ of the middle number and square it, you get the last number. 6 divided by 2 is 3, and 3 2 is 9. When this happens you have a special product. This problem factors into

Two terms must be squares, such as A² and B². Both of these terms will be positive. The remaining term must be 2AB or the opposite -2AB Factoring a Perfect-Square Trinomial A² + 2AB + B² = (A + B)² A² - 2AB + B² = (A - B)²

t² + 10t Check if the first and third terms are both perfect squares with positive coefficients. 2.If this is the case, identify a and b, and determine if the middle term equals 2ab The first term is a perfect square: t² = (t)² The third term is a perfect square: 25 = (5)² The middle term is twice the product of t and 5: 2(t)(5) t² + 10t +25 Perfect square trinomial

Determine whether each polynomial is a perfect-square trinomial Two of the terms in this expression are squares x² and 64. Twice the product of the square roots is 2·x·8, or 16x is a perfect square Only one term, 16 is a square (3x² is not a square because 3 is not a perfect square: 4x is not a square because x is not a square. is not a perfect square

t² + 4t +1 1.Check if the first and third terms are both perfect squares with positive coefficients. 2.If this is the case, identify a and b, and determine if the middle term equals 2ab The first term is a perfect square: t² = (t)² The third term is a perfect square: 1 = (1)² The middle term is not twice the product of t and 1: 2(t)(1) t² + 4t + 1 Is not perfect square trinomial

Determine whether each polynomial is a perfect-square trinomial Only one term, 16 is a square (3x² is not a square because 3 is not a perfect square: 4x is not a square because x is not a square. is not a perfect square

Determine whether this polynomial is a perfect-square trinomial First the polynomial should be put into descending order. Two of the terms, 100y² and 81, are squares. Twice the product of the square roots is 2(10y)(9), or 180y. The remaining term is the opposite of 180y. is a perfect square trinomial.

Remember: A perfect square trinomial is one that can be factored into two factors that match each other (and hence can be written as the factor squared). This is a perfect square trinomial because it factors into two factors that are the same and the middle term is twice the product of x and 6. It can be written as the factor squared. Notice that the first and last terms are perfect squares. The middle term comes from the outers and inners when Foilng. Since they match, it ends up double the product of the first and last term of the factor. Double the product of x and 6

The GCF is 1. The first and third terms are positive The first term is a perfect square: 25y² = (5y)² The third term is a perfect square: 4 = (2)² The middle term is twice the product of 5y and 2: 20y = 2(5y)(2) Factor as (5y - 2)²

y² - 25 The binomial is a difference of squares. = (y)² - (5)² Write in the form: a² - b², where a = y, b = 5. = (y + 5)(y – 5) Factor as (a + b)(a – b)

When you see two terms, look for the difference of squares. Is the first term something squared? Is the second term something squared but with a minus sign (the difference)? The difference of squares factors into conjugate pairs! difference rhyme for the day A conjugate pair is a set of factors that look the same but one has a + and one has a – between the terms. {

Look for something in common (there is a 5) Two terms left----is it the difference of squares? Yes---so factor into conjugate pairs. Factor Completely:

Look for something in common There is a 2p in each term Three terms left---try trinomial factoring "unFoiing" Check by FOILing and then distributing 2p through

Factoring a Sum and Difference of Cubes Sum of Cubes: a³+ b³ = (a + b)(a² -ab +b²) Difference of Cubes: a³ - b³ = (a - b)(a² +ab +b²)

= (x + 2) ( (x)² - (x)(2) + (2)²) x³ + 8 = (x)³ + (2) ³ x³ and 8 are perfect cubes -(x)(2) Square the first term of the binomial Product of terms in the binomial. Change the sign 2 x (x + 2) The factored form is the product of a binomial and a trinomial. The first and third terms in the trinomial are the squares of the terms within the binomial factor. Without regard to signs, the middle term in the trinomial is the product of terms in the binomial factor. Square the last term of the binomial.

If it's not the difference of squares, see if it is sum or difference of cubes. Is the first term something cubed (to the third power)? Is the second term something cubed? can be sum or difference here You must just memorize the steps to factor cubes. You should try multiplying them out again to assure yourself that it works. The first factor comes from what was cubed. square the first term multiply together but change signsquare the last term

What cubed gives the first term? What cubed gives the second term? Let's try one more: Try to memorize the steps to get the second factor: First term squared---multiply together & change sign- --last term squared The first factor comes from what was cubed. square the first term multiply together but change signsquare the last term