Question

using factor theorem factorise the polynomial x^4 -2x^3-7x^2+8x+12

Answer 1

Answer:

−2,−1,2,3.

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.

Answer 2

Answer:

Answer:

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

=x^4-3x^3+x^3-3x^2-4x^2+12x-4x+12

=x^3 (x-3)+x^2 (x-3)-4x (x-3)-4 (x-3)

=(x-3)(x^3+x^2-4x-4)

=(x-3)(x^3+2x^2-x^2-2x-2x-4)

=(x-3)[x^2 (x+2)-x (x+2)-2 (x+2)]

=(x-3)(x+2)(x^2-x-2)

=(x-3)(x+2)(x^2+x-2x-2)

=(x-3)(x+2)[x (x+1)-2 (x+1)]

=(x-3)(x+2)(x+1)(x-2)

This is your required answer

Step-by-step explanation: