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
Solution 1 When setting up the synthetic division tableau, we need to enter 0 for the coefficient of x in the dividend Doing so
AMSG 11 Remainder and Factor Theorem pdf
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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
6 The factor theorem
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As an example, consider the implication: 1) If no one is at home, then the machine answers the phone The converse of this is: 2) If the machine answers the
2 3 Factor and remainder theorems
www surrey ac uk/sites/default/files/2021-07/2 3-factor-and-remainder-theorems pdf
There are general algebraic solutions to cubic and quartic polynomial equations (analogous to the quadratic formula) Page 9 Some useful identities Page 10
5 1 The Remainder and Factor Theorems; Synthetic Division
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Example 5: Use both long and short (synthetic) division to find the quotient and (Zero = Root = Solution = x-intercept (if the zero is a real number))
MATHEMATICS SUPPORT CENTRE Title: Remainder Theorem and
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and factor theorems to find factors of polynomials Examples 1 Using previous example (Answers: yes, yes, no, yes, no) Example Factorise
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where cd = b A1ft: Correct factors for their 3-term quadratic followed by a solution (at least one value, which might be incorrect)
4 2 8 - The Factor Theorem - Scoilnet
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4 2 8 - The Factor Theorem 4 2 - Algebra - Solving Equations Leaving Certificate Mathematics Higher Level ONLY 4 2 - Algebra - Solving Equations
Apply the Remainder and Factor Theorems
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Given one solution of a polynomial equation, find the other solutions See Example 6 AVOID ERRORS The remainder after using synthetic division
AQA Core 1 Polynomials Section 2: The factor and remainder
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13 nov 2013 For example, for the cubic function g( ) ( 1)( 2)( 3) If there is an integer solution x = a, then by the factor theorem
Section 3 4 Factor Theorem and Remainder Theorem
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To find that "something," we can use polynomial division Example 1 Divide 14 5 4 2 3
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4.2.8 - The Factor Theorem
4.2 - Algebra - Solving Equations
Leaving Certi cate 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