factorise x2+6x+8 by factor theorum method Share with your friends Share 0 Varun.Rawat answered this The given polynomial is,px =x2 + 6x + 8The constant term in px is 8 and its factors are: ±1, ±2, ±4, ±8Put x = -2 in px, we getp-2 = -22 + 6-2 + 8 = 4 - 12 + 8 = 0⇒ x + 2 is a factor of px.Put x = -4 in px, we getp-4 = -42 +6 -4 + 8 = 16 - 24 + 8 = 0⇒ x +4 is a factor of pxSince px is a polynomial of degree 2, So it can't have more than 2 linear factors.Now,px = kx+2 x+4, k is an arbitary constant⇒ x2 + 6x+ 8 = k x+2 x+4put x = 0, we get 8 = k 2 4⇒ k = 1So, x2 + 6x+ 8 = x+2 x+4 1 View Full Answer Rishikesh Udhav Pandey answered this = X2 + 6x +8 = 0 = X2 + 4x + 2x + 8 = 0 = (X2 + 4x) + ( 2x + 8) = 0 = X (x + 4) + 2 (x + 4) = 0 = (x+ 2) ( x +4) Therefore, ( x + 2) ( x + 4) are the factors of x2 + 6x + 8. 0
