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
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
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,
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
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
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
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 - 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
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 – 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
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_628480.pdf
4.2.8 - The Factor Theorem
4.2 - Algebra - Solving Equations
Leaving Certicate Mathematics
Higher Level ONLY
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 1 / 5
The Factor Theorem
(x a) is a factor of the polynomialf(x) if and only iff(a) = 0.4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 2 / 5
The Factor Theorem
(x a) is a factor of the polynomialf(x) if and only iff(a) = 0.4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 2 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(i) x a=x 2)a= 2 )f(2)= 5(2
3) 14(22) + 12(2) 3= 40 56 + 24 3= 5
6= 0)(x 2)not a factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 3 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(ii) x a=x 1)a= 1 )f(1)= 5(1
3) 14(12) + 12(1) 3= 5 14 + 12 3= 0
)(x 1)isa factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 4 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(ii) x a=x 1)a= 1 )f(1)= 5(1
3) 14(12) + 12(1) 3= 5 14 + 12 3= 0
)(x 1)isa factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 4 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(ii) x a=x 1)a= 1 )f(1)= 5(1
3) 14(12) + 12(1) 3= 5 14 + 12 3= 0
)(x 1)isa factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 4 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(ii) x a=x 1)a= 1 )f(1)= 5(1
3) 14(12) + 12(1) 3= 5 14 + 12 3= 0
)(x 1)isa factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 4 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(ii) x a=x 1)a= 1 )f(1)= 5(1
3) 14(12) + 12(1) 3= 5 14 + 12 3= 0
)(x 1)isa factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 4 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(ii) x a=x 1)a= 1 )f(1)= 5(1
3) 14(12) + 12(1) 3= 5 14 + 12 3= 0
)(x 1)isa factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 4 / 5
Example 1
Q.Supposef(x) = 5x3 14x2+ 12x 3.
(i) Is (x 2) a factor? (ii) Is (x 1) a factor?(ii) x a=x 1)a= 1 )f(1)= 5(1
3) 14(12) + 12(1) 3= 5 14 + 12 3= 0
)(x 1)isa factor4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 4 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5
Example 2
Q.Findpif (x+ 3) is a factor off(x) = 4x3+ 21x2+px+ 12.Answer: x a=x+ 3)a= 3)f( 3)= 0
4( 3)3+ 21( 3)2+p( 3) + 12= 0
108 + 189 3p+ 12= 0 3p+ 93= 0 3p= 93)p= 31
4.2 - Algebra - Solving Equations4.2.8 - The Factor TheoremHigher Level ONLY 5 / 5