Factor Theorem
Let p(x) be a polynomial of degree n ≥ 1
and a be any real number. Then by factor theorem, x – a
is a factor of p(x), if p(a) = 0 and p(a)
= 0, if x – a is a factor of p(x).
FACTORISATION OF THE QUADRATIC POLYNOMIAL OF TYPE x2 + lx + m:
FACTORISATION OF THE QUADRATIC POLYNOMIAL OF TYPE x2 + lx + m:
For a quadratic polynomial of type x2
+ lx + m, where l, m are constants, first we split
the middle term lx as ax + bx so that ab =
m, then we get
x2 + lx + m | = x2 + ax + bx + ab |
= x(x + a) + b(x + a) | |
= (x + a)(x + b) |
FACTORISATION OF THE QUADRATIC POLYNOMIAL OF TYPE ax2 + bx + c:
For a quadratic polynomial of type axx2
+ bx + c, where a ≠ 0 and a, b, care
constants, first we assume that (px + q) and (rx + s) are its factors.
Then we get,
ax2 + bx + c | = (px + q) (rx + s) |
= prx2 + (ps + qr)x + qs |
Now comparing the coefficients of x2 on both sides we get,
a = pr
and comparing the coefficients of x on both sides we get,
b = ps + qr
Comparing the constant terms we get,
c = qs
a = pr
and comparing the coefficients of x on both sides we get,
b = ps + qr
Comparing the constant terms we get,
c = qs
Hence this implies that b is a sum of two numbers ps
and qr, and the product of these two numbers (ps)(qr) = (pr)(qs)
= ac.
Hence in order to factorise ax2 + bx + c, we have to write b as the sum of two numbers whose product is ac.
Hence in order to factorise ax2 + bx + c, we have to write b as the sum of two numbers whose product is ac.
Example 1:
Using the factor theorem determine whether x + 1 is a factor of
the polynomial 2x3
+ x2 – 2x - 1
or not.
Solution:
Let p(x) = 2x3 +
x2 – 2x – 1
Now x + 1 = 0 x = -1
Hence by factor theorem, x + 1 is a factor of p(x).
Now x + 1 = 0 x = -1
Hence p(-1) | = 2(-1)3 + (-1)2 – 2(-1) – 1 |
= -2 + 1 + 2 – 1 | |
= 0 |
Hence by factor theorem, x + 1 is a factor of p(x).
Example 2:
Factorise x2 – 5x
+ 6.
Solution:
Splitting the middle term we get:
Hence by factor theorem, x + 1 is a factor of p(x).
x2 - 5x + 6 | = x2 – 3x – 2x + 6 |
= x(x – 3) – 2(x – 3) | |
= (x- 3)(x – 2) (Answer) |
Hence by factor theorem, x + 1 is a factor of p(x).
Example 3:
Factorise 12x2 –
7x + 1.
Solution:
First we find the product of 12 so that the sum of the two numbers is 7. i.e., 12
= 4×3 (also 4 + 3 = 7)
Hence 12x2 – 7x + 1 | = 12x2 – 4x – 3x + 1 |
= 4x(3x – 1) -1(3x – 1) | |
= (3x – 1)(4x – 1) (Answer) |
Example 4:
Find the values of p and q so that x + 2 and x
- 1 are factors of the polynomial x3
+ 10x2 + px + q.
Solution:
Let g(x) = x3 +
10x2 + px + q
Now given that x + 2 is a factor of g(x).
By factor theorem if x + 2 is a factor of g(x), then g(-2) = 0.
Again if x – 1 is a factor of g(x), then g(-2) = 0.
g(1) = 0
Adding (1) and (2) we get,
3p = 21
=> p = 7
Putting the value of p = 7 in (2) we get,
7 + q = -11
=> q = -18
Hence p = 7 and q = -18 (Answer)
Now given that x + 2 is a factor of g(x).
By factor theorem if x + 2 is a factor of g(x), then g(-2) = 0.
Hence g(-2) | = (-2)3 + 10(-2)2 + p(-2) + q = 0 |
= -8 + 40 -2p + q = 0 | |
= 2p – q = 32 (1) |
Again if x – 1 is a factor of g(x), then g(-2) = 0.
g(1) = 0
Hence | (1)3 + 10(1)2 + p(1) + q = 0 |
=> 1 + 10 + p + q = 0 | |
=> p + q = -11 (2) |
Adding (1) and (2) we get,
3p = 21
=> p = 7
Putting the value of p = 7 in (2) we get,
7 + q = -11
=> q = -18
Hence p = 7 and q = -18 (Answer)
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