Question

What is Factor Theorum​

Answer 1

Answer:

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem. The factor theorem states that a polynomial has a factor ..

Answer 2

Step-by-step explanation:

If p(x) is a polynomial of degree n > 1 and a is any real number, then

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).

Let’s look at an example to understand this theorem better.

Example:

Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6.

Solution: To begin with, we know that the zero of the polynomial (x + 2) is –2. Let p(x) = x3 + 3x2 + 5x + 6

Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0

According to the factor theorem, if p(a) = 0, then (x – a) is a factor of p(x). In this example, p(a) = p(- 2) = 0

Therefore, (x – a) = {x – (-2)} = (x + 2) is a factor of ‘x3 + 3x2 + 5x + 6’ or p(x).

Factorisation of polynomials

You can factorise polynomials by splitting the middle term as follows: to begin with, consider a polynomial ax2 + bx + c with factors (px + q) and (rx + s). Therefore, we have ax2 + bx + c = (px + q) (rx + s). So, ax2 + bx + c = prx2 + (ps + qr) x + qs

If we compare the coefficients of x2, we get a = pr. Also, on comparing the coefficients of x, we get b = ps + qr. Finally, on comparing the constants, we get c = qs. Hence, b is the sum of two numbers ‘ps’ and ‘qr’, whose product is (ps)(qr) = (pr)(qs) = ac.

Therefore, to factorise ax2 + bx + c, we have to write b as the sum of two numbers whose product is ‘ac’. Let’s look at an example to understand this clearly.

Example

Factorise 6×2 + 17x + 5 by splitting the middle term.

Solution 1 (By splitting method): As explained above, if we can find two numbers, ‘p’ and ‘q’ such that, p + q = 17 and pq = 6 x 5 = 30, then we can get the factors.

After looking at the factors of 30, we find that numbers ‘2’ and ‘15’ satisfy both the conditions, i.e. p + q = 2 + 15 = 17 and pq = 2 x 15 = 30. So,

6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5

= 6x2 + 2x + 15x + 5

= 2x(3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Therefore, the factors of (6x2 + 17x + 5) are (3x + 1) and (2x + 5) with a remainder, zero.