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Section 6.4. Factoring and Solving Polynomial Equations. Factoring Polynomial Expressions. Previously, you learned how to factor the following types of quadratic expressions. Type. Example. General trinomial. 2 x 2 – 5 x – 12 = (2 x +3)( x – 4). Perfect square trinomial.

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## Section 6.4

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**Section 6.4**Factoring and Solving Polynomial Equations**Factoring Polynomial Expressions**Previously, you learned how to factor the following types of quadratic expressions. Type Example General trinomial 2x2 – 5x – 12 = (2x +3)(x – 4) Perfect square trinomial x2 + 10x + 25 = (x + 5)2 Difference of two squares 4x2 – 9 = (2x + 3) (2x – 3) Common monomial factor 6x2 + 15x = 3x(2x + 5) In this lesson you will learn how to factor other types of polynomials.**Factoring Polynomial Expressions**SPECIAL FACTORING PATTERNS Sum of Two Cubes Example a3 + b3 = (a + b)(a2 – ab + b2) x3 + 8 = (x + 2)(x2 – 2x + 4) Difference of Two Cubes a3 – b3 = (a – b)(a2 + ab + b2) 8x3 – 1 = (2x – 1)(4x2 + 2x + 1)**Factoring the Sum or Difference of Cubes**Factor each polynomial. x3 + 27 SOLUTION x3 + 27 = x3 + 33 Sum of two cubes. = (x + 3)(x2 – 3x + 9)**Factoring the Sum or Difference of Cubes**Factor each polynomial. x3 + 27 16u5 – 250u2 SOLUTION x3 + 27 = x3 + 33 Sum of two cubes. = (x + 3)(x2 – 3x + 9) 16u5 – 250u2 = 2u2(8u3 – 125) Factor common monomial. = 2u2[(2u)3 – 53] Difference of two cubes = 2u2(2u – 5)(4u2 + 10u +25)**Factoring by Grouping**For some polynomials, you can factor by grouping pairs of terms that have a common monomial factor. The pattern for this is as follows. ra + rb + sa + sb = r(a + b) + s(a + b) = (r + s)(a + b)**Factoring by Grouping**For some polynomials, you can factor by grouping pairs of terms that have a common monomial factor. The pattern for this is as follows. ra + rb + sa + sb = r(a + b) + s(a + b) = (r + s)(a + b) Factor the polynomial x3 – 2x2 – 9x + 18. SOLUTION x3 – 2x2 – 9x + 18 = x2(x – 2) – 9(x – 2) Factor by grouping. = (x2 – 9)(x – 2) = (x+ 3) (x– 3)(x – 2) Difference of squares.**Factoring Polynomials in Quadratic Form**An expression of the form au2 + bu + c where u is any expression in x is said to be in quadratic form. These can sometimes be factored using techniques you already know. Factor each polynomial 81x4 – 16 SOLUTION 81x4 – 16 = (9x2)2 – 42 = (9x2 + 4)(9x2 – 4) = (9x2 + 4)(3x +2)(3x – 2)**Factoring Polynomials in Quadratic Form**An expression of the form au2 + bu + c where u is any expression in x is said to be in quadratic form. These can sometimes be factored using techniques you already know. Factor each polynomial 81x4 – 16 4x6 – 20x4 + 24x2 SOLUTION SOLUTION 81x4 – 16 = (9x2)2 – 42 4x6 – 20x4 + 24x2 = 4x2(x4 – 5x2 + 6) = (9x2 + 4)(9x2 – 4) = 4x2(x2 – 2)(x2 – 3) = (9x2 + 4)(3x +2)(3x – 2)**Solving Polynomial Equations by Factoring**x = 0, x = 3, x = – 3, x = –2, orx = 2 The solutions are 0, 3, – 3, –2, and 2. Check these in the original equation. Solve 2x5 + 24x = 14x3 byusing the zero product property. SOLUTION 2x5 + 24x = 14x3 Write original equation. 2x5 – 14x3+ 24x = 0 Rewrite in standard form. 2x(x4 – 7x2 + 12) = 0 Factor common monomial. 2x(x2 – 3)(x2 – 4) = 0 Factor trinomial. 2x(x2 – 3)(x + 2)(x – 2) = 0 Factor difference of squares. Zero product property**Solving a Polynomial Equation in Real Life**Archeology In 1980 archeologists at the ruins of Caesara discovered a huge hydraulic concrete block with a volume of 330 cubic yards. The block’s dimensions are x yards high by 13x – 11 yards long by 13x – 15 yards wide. What is the height? SOLUTION Volume =Height• Length • Width Verbal Model Volume = 330 (cubic yards) Labels Height = x (yards) Length = 13x – 11 (yards) Width = 13x – 15 (yards) Algebraic Model 330 = x (13x – 11)(13x – 5)**Solving a Polynomial Equation in Real Life**The only real solution is x = 2, so 13x – 11 = 15 and 13x – 15 = 11. The block is 2 yards high. The dimensions are 2 yards by 15 yards by 11 yards. ArcheologyIn 1980 archeologists at the ruins of Caesara discovered a huge hydraulic concrete block with a volume of 330 cubic yards. The block’s dimensions are x yards high by 13x – 11 yards long by 13x – 15 yards wide. What is the height? SOLUTION 330 = x(13x – 11)(13x – 5) 0 = 169x3 – 338x2 + 165x – 330 Write in standard form. 0 = 169x2(x – 2) + 165(x – 2) Factor by grouping. 0 = (169x2 + 165)(x – 2)**Solving Polynomial Equations by Factoring**x = 0, x = 2, x = – 2, x = –2, orx = 2 The solutions are 0, 2, – 2, –2, and 2. Check these in the original equation. Solve 2x5 - 12x3 = -16x byusing the zero product property. SOLUTION 2x5 -12x3 = -16x Write original equation. 2x5 – 12x3+ 16x = 0 Rewrite in standard form. 2x(x4 – 6x2 + 8) = 0 Factor common monomial. 2x(x2 – 2)(x2 – 4) = 0 Factor trinomial. 2x(x2 – 2)(x + 2)(x – 2) = 0 Factor difference of squares. Zero product property**Section 6.4: page 348 – 349**# 18 – 24 (÷3), 27 – 32, 33 – 84 (÷3) Assignment

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