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Published byMadlyn Hancock Modified over 9 years ago
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Chapter 6 Section 4: Factoring and Solving Polynomials Equations
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1. Factor out any common monomials
3 π₯ 2 β12 =3 (π₯ 2 β4) π₯ 3 β2 π₯ 4 +x =π₯(π₯ 2 β2 π₯ 3 +1) 6π₯ 7 β4 π₯ 2 =2 π₯ 2 (3π₯ 5 β2)
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SPECIAL FACTORING PATTERNS
NAME Β Difference of Two Squares Perfect Square Trinomial a2 + 2ab + b2 = (a + b)2 PATTERN x2 β y2 = (x + y)(x β y) x2 + 12x + 36 = (x + 6)2 EXAMPLE x2 β 9 = (x + 3)(xβ 3)
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2. Look for special patterns
= (x + 4)(xβ 4) x2 β 16 x2 + 14x + 49 = (x + 7)2 There are other special patterns that are also worth remembering
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SPECIAL FACTORING PATTERNS
NAME Β Difference of Two cubes Sum of two cubes x3 + y3 = (x + y)(x2 βxy+ y2) PATTERN x3 - y3 = (x - y)(x2 +xy+ y2) x3 + 8 = (x + 2)(x2 -2x+ 4) EXAMPLE 8x3 β 1 (2x - 1)(4x2 +2x+ 1)
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3.Factoring by grouping 1. Begin by factoring out the GCF.
1. 5x3+2x2-40x-16 None 2. Arrange the four terms so that the first two terms and the last two terms have common factors. 2. (5x3+2x2)+(-40x-16) 3. If the coefficient of the third term is negative, factor out a negative coefficient from the last two terms. 3. (5x3+2x2)-(40x+16) 4. Use the reverse of the distributive property to factor each group of two terms. 4. x2(5x+2)-8(5x+2) 5. Now factor the GCF from the result of step 4 as done in the previous section. 5. (5x+2)(x2-8)
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Factoring using quadratics
x4 + 3x2 -4 The following steps can be used to solve equations that are quadratic in form: 1. Let u equal a function of the original variable (normally the middle term) 2. Substitute u into the original equation so that it is in the form au2 + bu + c 3. Factor the quadratic equation using the methods learned earlier 4. Replace u with the expression of the original variable. 5. Factor again if necessary. 1. u=x2 2. u2+3u-4 3. (u+4)(u-1) 4. (x2+4)(x2-1) 5. (x2+4)(x-1)(x+1)
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Solving Polynomials Remember
Finding zeros, solutions, and roots are different ways of saying the same thing. So⦠After you factor the polynomial, set it equal to 0. Then solve the polynomial.
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Find the real-number solutions
x4 + 3x2 -4=0 (x2+4)(x-1)(x+1)=0 (x2+4=0, not real (x-1)=0 x=1 (x+1)=0 x=-1
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