If f(x) is a polynomial and f(a) = 0, then (x–a) is a factor of f(x) Proof of the factor theorem Let's start with an example Consider 4 8 5
The Factor Theorem: Suppose p is a nonzero polynomial It is important to note that it works only for these kinds of divisors 5 Also Example 3 2 1
Now consider another example of a cubic polynomial divided by a linear divisor From the above example, we can deduce that: 2 ? 3 + 4 + 5 =
1 Factorise polynomial expressions 2 3 2 Divide a polynomial by a linear or quadratic factor 2 3 3 Apply the remainder theorem 2 3 4 Apply the factor theorem
For example, the zeros of p are –3, 1, and 5, and the factors of p(x) are x + 3, x - 1, and x - 5 The following theorem generalizes this relationship
2 Remainder and Factor Theorems Interactive Mathematics Factor theorem state with proof examples and solutions factorise the Polynomials Maths Mutt
Example 8: 7 5 4 3)( 2 3 + ? + = x x x xf Find )4( ? f using (a) synthetic division (b) the Remainder Theorem Example 9: Solve the equation
4 2 - Algebra - Solving Equations 4 2 8 - The Factor Theorem Higher Level ONLY 1 / 5 Example 1 Q Suppose f (x)=5x3 - 14x2 + 12x - 3
24 fév 2015 · Use long division to determine the other factors Page 6 6 February 24, 2015 Example Five Factor fully
PROBLEM You Have the Right to the Remainder Theorem Chapter 5 Polynomial Expressions and Equations example are correct? - Long Division
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101353_628480.pdf
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