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Chapter 6 Section 4: Factoring and Solving Polynomials Equations. 1. Factor out any common monomials. SPECIAL FACTORING PATTERNS. PATTERN NAME. Difference of Two Squares. Perfect Square Trinomial. a 2 + 2ab + b 2 = (a + b) 2. PATTERN. x 2 – y 2 = (x + y)(x – y ). - PowerPoint PPT Presentation
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Chapter 6Section 4: Factoring and
Solving Polynomials Equations
1. Factor out any common monomials
¿3 (𝑥¿¿2−4 )¿
¿2 𝑥2(3𝑥¿¿5−2)¿
SPECIAL FACTORING PATTERNS
PATT
ERN
NAM
E PA
TTER
N
EXAM
PLE
Difference of Two Squares
x2 – y2 = (x + y)(x – y)
x2 – 9 = (x + 3)(x– 3)
x2 + 12x + 36 = (x + 6)2
a2 + 2ab + b2 = (a + b)2
Perfect Square Trinomial
2. Look for special patterns
x2 – 16 = (x + 4)(x– 4)
x2 + 14x + 49 = (x + 7)2
There are other special patterns that are also worth remembering
SPECIAL FACTORING PATTERNS
PATT
ERN
NAM
E PA
TTER
N
EXAM
PLE
Difference of Two cubes
x3 + y3 =(x + y)(x2 –xy+ y2)
8x3 – 1 (2x - 1)(4x2 +2x+ 1)
x3 + 8= (x + 2)(x2 -2x+ 4)
x3 - y3 =(x - y)(x2 +xy+ y2)
Sum of two cubes
3.Factoring by grouping1. Begin by factoring out the GCF.
1. 5x3+2x2-40x-16None
2. Arrange the four terms so that the first two terms and the last two terms have common factors.
3. If the coefficient of the third term is negative, factor out a negative coefficient from the last two terms.4. Use the reverse of the distributive property to factor each group of two terms.
5. Now factor the GCF from the result of step 4 as done in the previous section.
2. (5x3+2x2)+(-40x-16)
3. (5x3+2x2)-(40x+16)
4. x2(5x+2)-8(5x+2)
5. (5x+2)(x2-8)
Factoring using quadratics
The following steps can be used to solve equations that are quadratic in form:
x4 + 3x2 -4
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)
Solving PolynomialsRemember
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.
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