In this case, The Remainder Theorem tells us the remainder when p(x) is divided by (x - c), namely p(c), is 0, which means (x - c) is a factor of p What we
What the theorem says, roughly speaking, is that if you have a zero of a polynomial, then you have a factor 6 factor theorem
24 fév 2015 · The Factor Theorem A special case of the remainder theorem The factor theorem states: • When a polynomial P(x) is evaluated at x = b and
Factor Theorem: The value a is a root of the polynomial p(x) if and only if (x?a) is a factor of p(x) Proof: 1 (=?) Assume that a is a root of the
(a) Use the factor theorem to show that (x + 4) is a factor of f (x) Using the remainder theorem, or otherwise, find the remainder when f(x) is divided
4 2 8 - The Factor Theorem 4 2 - Algebra - Solving Equations Leaving Certificate Mathematics Higher Level ONLY 4 2 - Algebra - Solving Equations
Descartes' Factor Theorem Drew Armstrong Descartes' La Géométrie (1637) is the oldest work of mathematics that makes sense to our modern eyes,
A SHORT PROOF OF THE FACTOR THEOREM FOR FINITE GRAPHS W T TUTTE We define a graph as a set V of objects called vertices together with a set E of
In this section, you will learn how to determine the factors of a polynomial function of degree 3 or greater Part 1: Remainder Theorem Refresher
![[PDF] 428 - The Factor Theorem - Scoilnet](https://pdfprof.com/EN_PDFV2/Docs/PDF_6/101352_628480.pdf.jpg "[PDF] 428 - The Factor Theorem - Scoilnet")
101352_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