Question

Factorise X^4-2x^3-7x^2+8x+12

Answer 1

Answer:

−2,−1,2,3.

Step-by-step explanation:

If we complete the square of x4−2x3, the last term

will be x2.

Rearranging the terms of P(x), we get,

P(x)=x4−2x3+x2−x2−7x2+8x+12,

=(x4−2x3+x2)−8x2+8x−−−−−−−−+12,

=(x2−x)2−8(x2−x)+12,

=y2−8y+12, say, where, y=x2−x,

=y2−6y−−−−−−−−2y+12−−−−−−−,

=y(y−6)−2(y−6),

=(y−6)(y−2),

={(x2−x)−6}{(x2−x)−2}..........[∵,y=x2−x],

={x2−3x−−−−−−−+2x−6−−−−−−}{x2−2x−−−−−−−+x−2−−−−−},

={x(x−3)+2(x−3)}{x(x−2)+1(x−2)},

⇒P(x)=(x−3)(x+2)(x−2)(x+1).

Hence, the zeroes of P(x) are, −2,−1,2,3.