Question

12.Solve (x^3-2x^2 +3x–4)(x–1)–(2x–3)(x2 –x+1)

Answer 1

Answer:

(x³-2x² +3x–4)(x–1)–(2x–3)(x² –x+1) = x⁴ - 5x³ + 10x² - 12x + 4

Step-by-step explanation:

We need to solve (x³-2x² +3x–4)(x–1)–(2x–3)(x² –x+1)

We first open the brackets

= [ x³(x - 1) -2x²(x - 1) +3x(x - 1) – 4(x - 1) ] - [ 2x(x² – x + 1) -3(x² – x + 1)]

On multiplying, we get

(x⁴ - x³ - 2x³ + 2x² + 3x² - 3x -4x + 1 ) - (2x³ - 2x² + 2x - 3x² + 3x - 3)

While multiplying, remember that (-)*(-) = (+) and (-)*(+) = (+)

Now, adding all the like terms together in both the brackets, we get

(x⁴ - 3x³ + 5x² - 7x + 1 ) - (2x³ - 5x² + 5x - 3)

= (x⁴ - 3x³ + 5x² - 7x + 1  - 2x³ + 5x² - 5x + 3)

Since (-)*(-) = (+) and (-)*(+) = (+)

Now, again adding all the like terms together, we get,

x⁴ - 5x³ + 10x² - 12x + 4

Answer 2

(x³-2x²+3x-4)(x-1) - (2x-3)(x²-x+1)

= (x³-2x²+3x-4)(x) + (x³-2x²+3x-4)(-1) - [2x(x²-x+1) + (-3)(x²-x+1)]

= x⁴-2x³+3x²-4x -x³+2x²-3x+4 - (2x³-2x²+2x -3x²+3x-3)

= x⁴-2x³+3x²-4x -x³+2x²-3x+4 -2x³+2x²-2x+3x²-3x+3

= x⁴-2x³-x³-2x³+3x²+2x²+2x²+3x²-4x-3x-2x-3x+4+3

= x⁴-5x³+10x²-12x+7 Answer