Question

The G.C.D of $({x^2} - 3x + 2),({x^2} - 4x + 4)$ is
A. \[(x - 2)\]
B. \[(x - 2)(x - 1)\]
C. \[{(x - 2)^2}\]
D. \[{(x - 2)^3}(x - 1)\]

Answer 1

Hint: We factorise both the polynomials and write their factors and take the factors common from both polynomials as their G.C.D.
* G.C.D means the greatest common divisor. We can calculate G.C.D of polynomials by factoring them. We can find G.C.D of two or more than two polynomials.
* If we are given any polynomial of the form ${x^2} - (a + b)x + ab$, then we can factorise this in two linear factors that is, ${x^2} - (a + b)x + ab = (x - a)(x - b)$
where when we substitute \[x = a\] and \[x = b\] in the polynomial, we get the value of the polynomial as zero.

Complete step-by-step answer:
We factorise both the polynomials separately.
Factorise the polynomial $({x^2} - 3x + 2)$
We can see here \[a + b = 3,ab = 2\]
By hit and trial method we look for numbers having sum three and multiplication as two.
So we have \[a = 1,b = 2\]
So, $({x^2} - 3x + 2) = (x - 1)(x - 2)$ ... (1)
Now we factorise the polynomial $({x^2} - 4x + 4)$
We can see here \[a + b = 4,ab = 4\]
By hit and trial method we look for numbers having sum as four and multiplication as four.
So we have \[a = 2,b = 2\]
So, $({x^2} - 4x + 4) = (x - 2)(x - 2)$ ... (2)
Now we find the G.C.D of the given polynomials
From (1) and (2), we have
$({x^2} - 3x + 2) = (x - 1)(x - 2)$ and $({x^2} - 4x + 4) = (x - 2)(x - 2)$
The common factor of $({x^2} - 3x + 2)$ and $({x^2} - 4x + 4) = (x - 2)$
Therefore, G.C.D of these two polynomials $ = (x - 2)$
G.C.D of $({x^2} - 3x + 2)$ and $({x^2} - 4x + 4)$ is $(x - 2)$

So, the correct answer is “Option A”.

Note: Students should not try to solve for values of a and b by substitution as it will again give a quadratic equation to solve. Also, keep in mind the number of factors will never exceed the highest power of the polynomial. Also, students can always check if the factors are correct or not by substituting the values \[x = a,x = b\] in the polynomial, the polynomial turns out to be zero when we substitute the values \[x = a,x = b\].