can anyone explain Factorisation of polynomial by using Factor Theorem
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hey mate!!
by factorisation theroem we can find that one polynomial is a factor or not
let's take one as g(x) and other as p(x), so first you have to put g(x)=0 and find the value of x and put in p(x ) by remainder theorem and if the remainder is zero than just at last you have to add a line that by factor theorem g(x) is a factor p(x)
for example
Determine whether x + 1 is a factor of the following polynomials.
a) 3x4 + x3 – x2 + 3x + 2
b) x6 + 2x(x – 1) – 4
Solution:
a) Let f(x) = 3x4 + x3 – x2 + 3x + 2
f(–1) = 3(–1)4 + (–1)3 – (–1)2 +3(–1) + 2
= 3(1) + (–1) – 1 – 3 + 2 = 0
Therefore, x + 1 is a factor of f(x)
b) Let g(x) = x6 + 2x(x – 1) – 4
g(–1) = (–1)6 + 2(–1)( –2) –4 = 1
Therefore, x + 1 is not a factor of g(x)
I hope this helps you!!!
by factorisation theroem we can find that one polynomial is a factor or not
let's take one as g(x) and other as p(x), so first you have to put g(x)=0 and find the value of x and put in p(x ) by remainder theorem and if the remainder is zero than just at last you have to add a line that by factor theorem g(x) is a factor p(x)
for example
Determine whether x + 1 is a factor of the following polynomials.
a) 3x4 + x3 – x2 + 3x + 2
b) x6 + 2x(x – 1) – 4
Solution:
a) Let f(x) = 3x4 + x3 – x2 + 3x + 2
f(–1) = 3(–1)4 + (–1)3 – (–1)2 +3(–1) + 2
= 3(1) + (–1) – 1 – 3 + 2 = 0
Therefore, x + 1 is a factor of f(x)
b) Let g(x) = x6 + 2x(x – 1) – 4
g(–1) = (–1)6 + 2(–1)( –2) –4 = 1
Therefore, x + 1 is not a factor of g(x)
I hope this helps you!!!
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