SECTION 2.4 Factoring polynomials. Do-Now  Factor the following polynomials.  x 2 + 5x + 6  4x 2 – 1  Solve the following equations.  2x 2 – 28x.

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Presentation transcript:

SECTION 2.4 Factoring polynomials

Do-Now  Factor the following polynomials.  x 2 + 5x + 6  4x 2 – 1  Solve the following equations.  2x 2 – 28x + 98 = 0  6x 2 – x – 2 = 0

Factoring Completely  A polynomial is factored completely if it is written as a product of unfactorable polynomials with integer coefficients.  Example: 4x(x 2 – 9) is not factored completely because x 2 – 9 can be factored into (x + 3)(x – 3).

Factor the following  y 3 – 4y 2 – 12y  Now we are working with polynomials of degree greater than 2. What should our first step be?  Always look to see if you can factor out a common monomial.  Additional examples:  3x x x  5x 5 – 80x 3

Factoring in quadratic form  Factor this expression:  x 2 – 5x – 14  Now try factoring…..  x 6 – 5x 3 – 14  Your answer should look just like a factored quadratic equation, but with bigger exponents  Additional Examples:  10x 4 – 10  3x x x 2

Set up and equation and solve.  The volume of the box is 96cm 3.  Write an equation to represent this situation and solve for x.  What is preventing us from solving this equation.

Factor by Grouping  When all else fails, group the x 3 and x 2 together and the x and constant term together. Factor them separately and see if the result is an expression that has a common factor.  Examples:  x 3 + 5x 2 + 3x + 15  x 3 – 3x 2 + 4x – 12

Be careful when the middle sign is negative….  Factor:  27x x 2 – 3x – 5  When grouping, rewrite as…..  (27x x 2 ) – (3x + 5)  Factor:  x 3 – 7x 2 – 9x + 63

Set up and equation and solve.  The volume of the box is 96cm 3.  Write an equation to represent this situation and solve for x.  Now use your factoring by grouping skills to solve the equation.

Sum or Difference of Cubes

EXAMPLE 2 Factor the sum or difference of two cubes Factor the polynomial completely. a. x = (x + 4)(x 2 – 4x + 16) Sum of two cubes b. 16z 5 – 250z 2 Factor common monomial. = 2z 2 (2z) 3 – 5 3 Difference of two cubes = 2z 2 (2z – 5)(4z z + 25) = x = 2z 2 (8z 3 – 125)