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6 The factor theorem
6 The factor theorem www mast queensu ca/~peter/investigations/6factors pdf It's worth pointing out that cubic equations are not so easy to solve If the equation in Example 3 were quadratic, we could use the quadratic formula, but it's
23 Factor and remainder theorems
2 3 Factor and remainder theorems www surrey ac uk/sites/default/files/2021-07/2 3-factor-and-remainder-theorems pdf knowledge and skills in working with the factor and remainder theorems It 2 3 3 Apply the remainder theorem Example: Remainder theorem
32 The Factor Theorem and The Remainder Theorem
3 2 The Factor Theorem and The Remainder Theorem www shsu edu/~kws006/Precalculus/2 3_Zeroes_of_Polynomials_files/S 26Z 203 2 pdf Example 3 2 1 Use synthetic division to perform the following polynomial divisions Find the quotient and the remainder polynomials, then write the dividend,
AMSG11Remainder and Factor Theorempdf
AMSG 11 Remainder and Factor Theorem pdf irp-cdn multiscreensite com/f15f3f52/files/uploaded/AMSG 11 Remainder 20and 20Factor 20Theorem pdf For example, we may solve for x in the following equation as follows: Hence, x = ?3 or ?2 are solutions or roots of the quadratic equation A more general
22 - The Factor Theorem
2 2 - The Factor Theorem vanvelzermath weebly com/uploads/2/3/5/2/23525212/2 2_the_factor_theorem pdf 24 fév 2015 Use long division to determine the other factors Page 6 6 February 24, 2015 Example Five Factor fully
Factor Theorem
Factor Theorem mr-choi weebly com/uploads/1/7/0/5/17051620/2-2_-_factor_theorem pdf 2 2 - Factor Theorem Factor Theorem Example 1a: Use the factor theorem to determine which binomials are factors of the polynomial
428 - The Factor Theorem - Scoilnet
4 2 8 - The Factor Theorem - Scoilnet www scoilnet ie/uploads/resources/28744/28480 pdf Example 1 Q Suppose f (x)=5x3 - 14x2 + 12x - 3 (i) Is (x - 2) a factor? (ii) Is (x - 1) a factor? 4 2 - Algebra - Solving Equations 4 2 8 - The Factor
Factor Theorem - jongarvincom
Factor Theorem - jongarvin com jongarvin com/up/MHF4U/slides/factor_theorem_handout pdf Factor Theorem J Garvin Slide 1/14 polynomial equations & inequalities Factor Theorem Example Divide f (x) = x3 + 4x2 + x - 6 by x - 1
The Factor Theorem and a corollary of the - UMass Blogs
The Factor Theorem and a corollary of the - UMass Blogs blogs umass edu/math421-murray/files/2010/08/FactorTheoremEvaluated pdf 27 août 2010 the latter inequality says that the remainder r is less than the “divisor” b For example, if you use long division to divide 2356 by 14, you
L3 – 22 – Factor Theorem Lesson MHF4U - jensenmath
L3 – 2 2 – Factor Theorem Lesson MHF4U - jensenmath www jensenmath ca/s/22-ls-factor-theorem pdf a) Use the remainder theorem to determine the remainder when Example 1: Determine if ?3 and +2 are factors of ( ) = ? ? 14 + 24
51 The Remainder and Factor Theorems; Synthetic Division
5 1 The Remainder and Factor Theorems; Synthetic Division users math msu edu/users/bellro/mth103fa13/mth103fa13_chapter5 pdf use the factor theorem Example 1: Use long division to find the quotient and the remainder: 27 5593 ÷ Steps for Long Division:
99584_6factor_theorem_handout.pdf polynomial equations & inequalitiespolynomial equations & inequalities Factor TheoremExample
Factorf(x) =x3-3x2-50.Test values ofxto find a factor off(x).f(1) =-52, sox-1 is not a factor.f(2) =-54, sox-2 is not a factor.f(3) =-50, sox-3 is not a factor.f(4) =-34, sox-4 is not a factor.f(5) = 0, sox-5 is a factor.J. Garvin- F actorTheo rem
Slide 4/14MHF4U: Advanced Functions
Factor Theorem
J. Garvin
Slide 1/14polynomial equations & inequalities
Factor TheoremExample
Dividef(x) =x3+ 4x2+x-6 byx-1.1 4 1-6
11 5 6
1 5 6 0
Sof(x) = (x-1)(x2+ 5x+ 6), with no remainder.Therefore,x-1 is a factor off(x).J. Garvin- F actorTheo rem
Slide 2/14polynomial equations & inequalities
Factor Theorem
This application of the Remainder Theorem is called the Factor Theorem.Factor Theorem (FT) for Polynomials x-bis a factor of polynomialP(x) iffP(b) = 0, andax-b is a factor ofP(x) iffP?ba ?= 0.When used with synthetic (or long) division, the FT provides a method of factoring polynomials of any degree, assuming they are factorable.By determining values ofxsuch thatP(x) = 0, we can determine factors ofP(x), then perform a division to reduce the degree of the polynomial.J. Garvin- F actorTheo remSlide 3/14polynomial equations & inequalities
Factor TheoremExample
Factorf(x) =x3-3x2-50.Test values ofxto find a factor off(x).f(1) =-52, sox-1 is not a factor.f(2) =-54, sox-2 is not a factor.f(3) =-50, sox-3 is not a factor.f(4) =-34, sox-4 is not a factor.f(5) = 0, sox-5 is a factor.J. Garvin- F actorTheo rem
Slide 4/14polynomial equations & inequalities
Factor Theorem
Use synthetic division to determine the quotient when f(x) =x3-3x2-50 is divided byx-5.1-3 0-5055 10 50
1 2 10 0
Sof(x) = (x-5)(x2+ 2x+ 10).x
2+ 2x+ 10 is not factorable, sof(x) is fully factored.J. Garvin- F actorTheo rem
Slide 5/14polynomial equations & inequalities
Integral Zero Theorem
Guessing values forxmay take a long time, so there are some ways in which we can reduce the sample space.Consider a polynomial in both factored form, f(x) =k(x-r1)(x-r2)...(x-rn), and standard form,f(x) =anxn+an-1xn-1+...+a1x+a0.Note thata0=k×r1×r2×...×rn.Therefore, ifx-bis a factor off(x), it must be the case
thatbis a factor ofa0.In the previous example,x-5 was a factor off(x), and 5 isa factor of-50.x-3 was not a factor off(x), and 3 is not a factor of-50.However,x-2 was not a factor off(x), even though 2 is a
factor of-50.J. Garvin- F actorTheo remSlide 6/14
polynomial equations & inequalitiesIntegral Zero Theorem
This insight is known as theIntegral Zero Theorem.Integral Zero Theorem (IZT) for PolynomialsIfx-bis a factor of polynomialP(x), with integer
coefficients an leading coefficient 1, thenbis a factor of the constant term ofP(x).Factors may be either positive or negative. As its name suggests, the IZT provides a method of findingall zeroes that can be expressed as integers.Any non-integral solutions will not be found using the IZT,
but it provides a good starting point.J. Garvin- F actorTheo remSlide 7/14polynomial equations & inequalities
Integral Zero TheoremExample
Factorf(x) =x3+ 6x2-x-30.Factors of-30 are±1,±2,±3,±5,±6,±10,±15 and±30.Sincef(1) =-5,x-1 is not a factor off(x).Sincef(2) = 0,x-2 is a factor off(x).1 6-1-30
22 16 30
1 8 15 0
Therefore,f(x) = (x-2)(x2+ 8x+ 15).Factoring the simple trinomial,f(x) = (x-2)(x+ 3)(x+ 5).J. Garvin- F actorTheo rem
Slide 8/14polynomial equations & inequalities
Rational Zero Theorem
The IZT may be extended to handle rational numbers, ba .Rational Zero Theorem (RZT) for Polynomials IfP(x) is a polynomial with integer coefficients, and ifba is a rational zero ofP(x), thenbis a factor of the constant term ofP(x) andais a factor if the leading coefficient.Again, factors may be either positive or negative. While the RZT provides additional possible factors, it is generally it is a good idea to check for integers first.J. Garvin- F actorTheo remSlide 9/14polynomial equations & inequalities
Rational Zero TheoremExample
List all possible rational zeroes forf(x) = 2x3-5x+ 6.Factors of the constant term, 6, are±1,±2,±3 and±6.Factors of the leading coefficient, 2, are±1 and±2.By the RZT, the possible rational zeroes off(x) are±1,±2,
±3,±6,±12
and±32 .J. Garvin- F actorTheo remSlide 10/14polynomial equations & inequalities
Rational Zero TheoremExample
Determine the real roots off(x) = 2x3-3x2-10x+ 15.By the RZT, the possible rational zeroes off(x) are±1,±3,
±5,±15,±12
,±32 ,±52 and±152 .Sincef?32 ?= 0, 2x-3 is a factor off(x).2-3-10 15323 0-152 0-10 0Dividing the quotient by 2,f(x) = (2x-3)(x2-5).Sincex2-5 = (x-⎷5)(x+⎷5), the real roots off(x) are
32and±⎷5.