Factorising 6x^2-13x+6 using factor theorem

The factor theorem states that if a polynomial P(x) has a factor (x-a), then P(a)=0. Using this theorem, we can factorise the given polynomial 6x^2-13x+6 as follows:

Step 1: Find the factors of the leading coefficient (6) and the constant term (6). The factors of 6 are 1, 2, 3 and 6, while the factors of 6 are 1, 2, 3 and 6.

Step 2: Using the factors of the leading coefficient and the constant term, we can make a list of possible factors of the polynomial. In this case, the possible factors are:

(2x - 1)(3x - 6)
(2x - 2)(3x - 3)
(2x - 3)(3x - 2)
(2x - 6)(3x - 1)

Step 3: We can check which of these factors is correct by using the factor theorem. We substitute each factor into the polynomial and see if it equals zero. For example, if we substitute (2x - 1)(3x - 6) into the polynomial, we get:

P(2/3) = 6(2/3)^2 - 13(2/3) + 6 = 0

Since P(2/3) equals zero, we know that (2x - 1)(3x - 6) is a factor of the polynomial.

Step 4: Once we have found one factor of the polynomial, we can use long division or synthetic division to divide the polynomial by the factor and find the other factor. In this case, we can use long division to get:

6x^2 - 13x + 6 = (2x - 1)(3x - 6)
= 2x(3x - 6) - 1(3x - 6)
= (2x - 1)(3x - 6)

Therefore, the factorisation of 6x^2 - 13x + 6 using the factor theorem is (2x - 1)(3x - 6).

6x^2-13x 6 =6x^2-4x-9x 6 =2x(3x-2)-3(3x-2) =(2x-3)(3x-2)