Question

find GCDby factorization. x2-x-2,x2+x-6,x2-3x+2​

Answer 1

Solution :

1st term = x² - x - 2

 = x² - (2 - 1)x - 2

 = x² - 2x + x - 2

 = x (x - 2) + 1 (x - 2)

 = (x - 2) (x + 1)

2nd term = x² + x - 6

 = x² + (3 - 2)x - 6

 = x² + 3x - 2x - 6

 = x (x + 3) - 2 (x + 3)

 = (x + 3) (x - 2)

3rd term = x² - 3x + 2

 = x² - (1 + 2)x + 2

 = x² - x - 2x + 2

 = x (x - 1) - 2 (x - 1)

 = (x - 1) (x - 2)

In 1st, 2nd and 3rd term, we see that there is common factor (x - 2)

Therefore, the required gcd = (x - 2)

Answer 2

$Factorising\:First\:equation:x^{2} -x-2\\\\=x^{2} -2x+1x-2\\\\=x(x-2)+1(x-2)\\\\=(x+1)(x-2)\\\\Factorising\:second\:equation:x^{2} +x-6\\\\=x^{2} +3x-2x-6\\\\=x(x-2)+3(x-2)\\\\=(x+3) (x-2)\\\\Factorising\:third\:equation:x^{2} -3x+2\\\\=x^{2} -2x-1x+2\\\\=x(x-1)-2(x-1)\\\\=(x-1)(x-2)\\\\Common \:factor\:in\:all\:three\:equations:(x-2)\\\\Thus,\:the\:GCD\:is\:(x-2)$