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2.4 Factor and Solve Polynomial Equations

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## 2.4 Factor and Solve Polynomial Equations

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**2.4 Factor and Solve Polynomial Equations**p. 111 Name two special factoring patterns for cubes. Name three ways to factor a polynomial. What is the difference between factoring a polynomial and solving a polynomial?**Types of Factoring:**From Chapter 1 we did factoring of: • GCF : 6x2 + 15x = 3x (2x + 5) 1stongold card • PTS : x2 + 10x + 25 = (x + 5)2 patterns • DOS : 4x2 – 9 = (2x + 3)(2x – 3) difference of squares • Bustin’ da B = 2x2 – 5x – 12 = • (2x2 - 8x) + (3x – 12) = • 2x(x – 4) + 3(x – 4)= • (x – 4)(2x + 3)**Factor the polynomial completely.**a. x3 + 2x2 – 15x = x(x2 + 2x – 15) Factor common monomial. = x(x + 5)(x – 3) Factor trinomial. = 2y3(y2 – 9) b. 2y5 – 18y3 Factor common monomial. = 2y3(y + 3)(y – 3) Difference of two squares = 4z2(z2 – 4z + 4) c. 4z4 – 16z3 + 16z2 Factor common monomial. = 4z2(z – 2)2 Perfect square trinomial**Now we will use Sum of Cubes:**• a3 + b3 = (a + b)(a2 – ab + b2) • x3 + 8 = • (x)3 + (2)3 = • (x + 2)(x2 – 2x + 4)**Difference of Cubes**• a3 – b3 = (a – b)(a2 + ab + b2) • 8x3 – 1 = • (2x)3 – 13 = • (2x – 1)((2x)2 + 2x*1 + 12) • (2x – 1)(4x2 + 2x + 1)**When there are more than 3 terms – use GROUPING**• x3 – 2x2 – 9x + 18 = • (x3 – 2x2) + (-9x + 18) = Group in two’s • with a ‘+’ in the middle • x2(x – 2) - 9(x – 2) = GCF each group • (x – 2)(x2 – 9) = • (x – 2)(x + 3)(x – 3) Factor all that can be • factored**= 2z2 (2z)3 – 53**Factor the polynomial completely. = x3 + 43 a. x3 + 64 Sum of two cubes = (x + 4)(x2 – 4x + 16) = 2z2(8z3 – 125) b. 16z5 – 250z2 Factor common monomial. Difference of two cubes = 2z2(2z – 5)(4z2 + 10z + 25)**Factor the polynomial completely.**1. x3 – 7x2 + 10x SOLUTION x3 – 7x2 + 10x = x3 – 7x2 + 10x = x( x2 – 7x + 10) Factor common monomial. = x( x – 5 )( x – 2 ) Factor trinomial.**2. 3y5 – 75y3**SOLUTION Factor common monomial. 3y5 – 75y3 = 3y3 (y2 – 25) = 3y3 (y – 5)( y + 5 ) Difference of two squares**3. 16b5 + 686b2**SOLUTION = 2b2 (8b3 + 343) Factor common monomial. = 2b2 (2b + 7)(4b2 –14b + 49 ) Difference of two cubes 4. w3 – 27 SOLUTION w3 – 27 =w3 – (3)3 Difference of two cubes =(w – 3)(w2 + 3w + 9)**Factor by Grouping**Factor the polynomial x3 – 3x2 – 16x + 48 completely. = x2(x – 3) –16(x – 3) x3– 3x2– 16x + 48 Factor by grouping. = (x2 – 16)(x – 3) Distributive property = (x+ 4)(x – 4)(x – 3) Difference of two squares Can you factor this by using the box method—YES if there are 4 terms**Factoring in Quadratic form:**• 81x4 – 16 = • (9x2)2 – 42 = • (9x2 + 4)(9x2 – 4)= Can anything be • factored still??? • (9x2 + 4)(3x – 2)(3x +2) • Keep factoring ‘till you can’t factor any more!!**You try this one!**• 4x6 – 20x4 + 24x2 = • 4x2 (x4 - 5x2 +6) = • 4x2 (x2 – 2)(x2 – 3)**In Chapter 1, we used the zero property. (when multiplying 2**numbers together to get 0 – one must be zero)The also works with higher degree polynomials**Solve:**• 2x5 + 24x = 14x3 • 2x5 - 14x3 + 24x = 0 Put in standard form • 2x (x4 – 7x2 +12) = 0 GCF • 2x (x2 – 3)(x2 – 4) = 0 Bustin’ da ‘b’ • 2x (x2 – 3)(x + 2)(x – 2) = 0 Factor • everything • 2x=0 x2-3=0 x+2=0 x-2=0 set all • factors to 0 • X=0 x=±√3 x=-2 x=2**Now, you try one!**• 2y5 – 18y = 0 • Y=0 y=±√3 y=±i√3**City Park**You are designing a marble basin that will hold a fountain for a city park. The basin’s sides and bottom should be 1 foot thick. Its outer length should be twice its outer width and outer height. What should the outer dimensions of the basin be if it is to hold 36 cubic feet of water?**The only real solution is x = 4. The basin is 8ft long, 4ft**wide, and 4ft high. ANSWER SOLUTION 36 = (2x – 2)(x – 2)(x – 1) Write equation. 0 = 2x3 – 8x2 + 10x – 40 Write in standard form. 0 = 2x2(x – 4) + 10(x – 4) Factor by grouping. 0 = (2x2 + 10)(x – 4) Distributive property**Name two special factoring patterns for cubes.**Sum of two cubes and difference of two cubes. • Name three ways to factor a polynomial. Factor by sum or difference of cubes, by grouping or when in quadratic form. • What is the difference between factoring a polynomial and solving a polynomial? Solving takes the problem one step further than factoring.**2.4 Assignment**Page 111, 3 – 48 every third problem