 Polynomials Lesson 5 Factoring Special Polynomials.

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Presentation transcript:

 Polynomials Lesson 5 Factoring Special Polynomials

Todays Objectives  Students will be able to demonstrate an understanding of the factoring of polynomial expressions, including:  Factor a polynomial that is a difference of squares, and explain why it is a special case of trinomial factoring where b = 0  Identify and explain errors in a solution for a polynomial expression

Factoring Special Polynomials  Today we will look at factoring two types of special polynomials  Perfect square trinomials  Difference of squares

Factoring Special Polynomials  Consider a square with side length a + b: ab a b It’s area is: (a + b) 2 = (a + b)(a + b) =a(a + b) + b(a + b) =a 2 + ab + ab + b 2 =a 2 + 2ab + b 2

Perfect Square Trinomials  We say that a 2 + 2ab + b 2 is a perfect square trinomial:  a 2 is the square of the first term in the binomial  2ab is twice the product of the first and second terms in the binomial  b 2 is the square of the second term in the binomial  When we use algebra tiles to represent a perfect square trinomial, the tiles will form a square shape.

Perfect Square Trinomials  2 forms of perfect square trinomials:  (a – b) 2 = a 2 – 2ab + b 2  (a + b) 2 = a 2 + 2ab + b 2  We can use these patterns to factor perfect square trinomials.

Example  Factor the trinomial using your algebra tiles: 4x x + 9  Solution:  Arrange the algebra tiles to form a square. The side lengths will be equal to 2x + 3. So we can see that this trinomial is a perfect square trinomial with factors (2x + 3) 2.  Check:  (2x + 3)(2x + 3) = 2x(2x + 3) + 3(2x + 3)=4x 2 + 6x + 6x + 9 = 4x x + 9  The result is the same as our original trinomial, so the factors are correct

Example  Factor the trinomial 4 – 20x + 25x 2  Solution:  The first term is a perfect square (4 = 2 x 2)  The third term is a perfect square (5x)(5x) = 25x 2  The second term is twice the product of 5x and 2 (10x)(2) = 20x  Since the 2 nd term is negative, the operations in the binomial factors must be subtraction.  So, the trinomial is a perfect square with factors: (2 – 5x)(2 – 5x) or (2-5x) 2.

Example  Factor the trinomial 16 – 56x + 49x 2  Solution:  (4 – 7x) 2

Difference of Squares  Another example of a special polynomial is a difference of squares. A difference of squares is a binomial of the form a 2 – b 2. We can think of this as a trinomial with a middle term of zero. For example, we could write the perfect square (x 2 – 25) as the trinomial (x 2 – 0x – 25).  This is a perfect square because x 2 = (x)(x), and 25 = (5)(5). Any subtraction expression is known as a difference. Therefor, this is a difference of squares.

Factoring a Difference of Squares  To factor this “trinomial”, we should find two integers whose product is -25, and whose sum is 0. These two integers are 5 and -5.  So, x 2 – 25 = (x + 5)(x – 5).  This pattern is true for any difference of squares.

Example  Factor the difference of squares 25 – 36x 2  Solution:  Write each term as a perfect square.  25 – 36x 2 = (5) 2 – (6x) 2 = (5 + 6x)(5 – 6x)

Example  Factor 5x 4 – 80y 4  Solution:  As written in this example, each term is not a perfect square, but we can remove a common factor of 5.  = 5(x 4 – 16y 4 )  = 5[(x 2 ) 2 – (4y 2 ) 2 ]  = 5(x 2 – 4y 2 )(x 2 + 4y 2 )  The first binomial is also a difference of squares  = 5(x + 2y)(x – 2y)(x 2 + 4y 2 )

Example  Factor the following:  81m 2 – 49  Solution: (9m+7)(9m-7)  162v 4 – 2w 4  Solution: 2(81v 4 -w 4 ) = (2)(9v 2 +w 2 )(9v 2 -w 2 )  =(2)(9v 2 +w 2 )(3v+w)(3v-w)