There, we explained the terms Factors or Divisors,
Factorization of Polynomials, Prime Polynomial etc.
That knowledge is a prerequisite here.
Here, we factorize higher degree ( > 2 ) polynomials using
Factor Theorem which is based on Remainder Theorem.
Remainder Theorem
If a Polynomial in x, say f(x) is divided by (x - a),
then the remainder is f(a).
Proof :
We know Dividend = Divisor x Quotient + Remainder
Let q(x) be the Quotient and r be the Remainder
when f(x) is divided by (x - a).
Then we have f(x) = (x - a) x q(x) + r where in the degree of q(x) is
one less than that of f(x) and r is a constant.
Since the equation is true for all real values of x,
putting x = a in the equation, we get
f(a) = (a - a) x q(a) + r = 0 + r = r ⇒ r = f(a).
Thus, when f(x) is divided by (x - a), the remainder is f(a). (Proved.)
Note:
If f(x) is divided by (ax + b), then the remainder is f(-b ⁄a)
The proof for this is similar to the one given above.
Example on Remainder Theorem : Factoring Polynomials
Find the remainder when 3x3 - 2x2 + x + 2
is divided by (i) x - 1 (ii) x + 2 (iii) 2x -1 (iv) 3x + 2
Solution:
Let f(x) = 3x3 - 2x2 + x + 2
(i) when f(x) is divided by (x - 1),
The remainder = f(1) = 3(1)3 - 2(1)2 + (1) + 2 = 3 - 2 +1 + 2 = 4. Ans.
(ii) when f(x) is divided by (x + 2),
The remainder = f(-2) = 3(-2)3 - 2(-2)2 + (-2) + 2 = 3(-8) - 2(4) - 2 + 2 = -32. Ans.
(iii) when f(x) is divided by (2x -1),
The remainder = f(1⁄2) = 3(1⁄2)3 - 2(1⁄2)2 + (1⁄2) + 2
= 3(1⁄8) - 2(1⁄4) + 1⁄2 + 2 = (3 - 2 x 2 + 1 x 4 + 2 x 8)⁄8 = 19⁄8 Ans.
(iv) when f(x) is divided by (3x + 2),
The remainder = f(-2⁄3) = 3(-2⁄3)3 - 2(-2⁄3)2 + (-2⁄3) + 2
= 3(-8⁄27) - 2(4⁄9) - 2⁄3 + 2 = (3 x -8 - 2 x 4 x 3 - 2 x 9 + 2 x 27)⁄27 = -12⁄27 = -4⁄9 Ans.
An important application of Remainder Theorem is the
Factor Theorem which is applied in Factoring polynomials.
Factor Theorem : Factoring Polynomials
We know that, if the remainder is zero, when a Polynomial f(x) is divided by (x - a),
then (x - a) is a factor of f(x).
But, as per remainder theorem, remainder = f(a).
∴ If f(a) = 0, then (x - a) is a factor of f(x).
Also If (x - a) is a factor of f(x), then f(a) = 0.
This is Factor theorem.
(x - a) is a factor of f(x) ⇔ f(a) = 0
Factor Theorem is very useful in Factoring
Polynomials of higher (than 2) degree .
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Method of Factoring Polynomials using Factor Theorem :
The method of Factoring Polynomials
involves the following steps.
STEP 1: Let the Polynomial be f(x). Check whether (x- a) is a factor of f(x) or not.
To check whether (x - a) is a factor of f(x) or not, we have to check
whether f(a) is zero or not. Choosing values of a is done by trial and error.
To facilitate, reducing the number of trials, we take the constant term
of the Polynomial, f(x), find its factors, put + and - to the factors and take
those values for a and check with them, if f(a) is zero or not. This is
done till we find an a such that f(a) is zero.
STEP 2: After deciding (x - a) is a factor, we divide the Polynomial f(x) by
(x - a) to get the Quotient. This division is better done by Synthetic Division
which is explained in Solved Example 1 of Set 3 below.
STEP 3: we, then Factorize the Quotient for further factors. This Factorizing
is done either by Factor theorem (Repeat from Step 1) or by other methods
depending upon the degree of the Quotient.
The method of Factoring Polynomials will be clear by the following set of Solved Examples.
Example 1 of Factoring Polynomials
Solve the following problem on Factoring Polynomials
If a and b are unequal and x2 + ax + b and x2 + bx + a have a common factor,
show that a + b + 1 = 0
Solution to Example 10 of Factoring Polynomials
Let f(x) = x2 + ax + b; and p(x) = x2 + bx + a
Let (x - k) be the common factor of f(x) and p(x).
⇒ f(k) = 0 and p(k) = 0
f(k) = 0 ⇒ k2 + ak + b = 0............(i)
p(k) = 0 ⇒ k2 + bk + a = 0............(ii)
(i) - (ii) gives (k2 + ak + b) - (k2 + bk + a) = 0 - 0 = 0
⇒ ak + b - bk - a = 0 ⇒ k(a - b) -1(a - b) = 0 ⇒ (a - b)(k- 1) = 0
⇒ ( (k- 1) = 0 [ Since by data a and b are unequal ⇒ (a - b) ≠ 0 ]
⇒ k = 1.
Substituting this value of k in (i), we get
(1)2 + a(1) + b = 0 ⇒ a + b + 1 = 0
Thus,
If a and b are unequal and x2 + ax + b and x2 + bx + a have a common factor,
we could show that a + b + 1 = 0 using Factor Theorem in Factoring polynomials.
Solve the Following Problems on Factoring polynomials :
If f(x) = x2 + 5x + a and p(x) = x2 + 3x + b have a common factorthen (i) find the common factor (ii) show that (a - b)2 = 2(3a - 5b)
For Answer See at the bottom of the Page.
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