Question

factorise (x² - x)² -8 (x²-x)+ 12​

Answer 1

Answer:

x⁴+2x³-7x²-8x+12

Step-by-step explanation:

Consider the following polynomial.

f(x)=x⁴+2x³-7x²-8x+12

The constant term is 12

Therefore the possible zeroes are ±1,± 2,± 3, ±4 ±6, ± 12.

Put x=1 in f(x)

f(1)=(1)⁴+2(1)³-7(1)²-8(1)+12

= 15-15

= 0

= (1)+2 (1) 7 (1) 8 (1) + 12

= 1+2-7-8 +12

Since f(1)=0 , therefore by factor theorem (x-1) is a factor of f(x)

Put x=-3 in f(x)

f(-3)=(-3)⁴+2(-3)³-7(-3)²-8(-3)+12

= 81 + 2 (-27) - 7 (9) + 24 + 12

= 81 54 63 + 24 + 12

-

= 117 117

= 0

Since f(-3)=0 , therefore by factor theorem (x+3) is a factor of f (x)

Put x=2 in f(x)

f(2)=(2)⁴+2(2)³-7(2)²-8(2)+12

= 16+2 (8) 7 (4) - 16 +12

= 16+16 28 16 +12

-

= 44-44

= 0

Since f(2)=0 , therefore by factor theorem (x-2) is a factor of f(x)

Put x=-2 in f(x)

f(-2)=(-2)⁴+2(-2)³-7(-2)²-8(-2)+12

= 16+ 2 (-8) - 7 (4) + 16 +12

= 16-16-28 + 16 + 12

= -28 +28

= 0

Since f(-2)=0 , therefore by factor theorem (x+2) is a factor of f(x)

Hence the required factors of given polynomial are

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