Question

factorise: -x^2y-x+3xy+3
please answer someone correctly. i will mark as brainliest.​

Answer 1

Answer:

- x²y-x + 3xy + 3

= 3 -x + 3xy - x²y

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

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

Answer 2

ANSWER:

(3 - x)(xy + 1)

STEP-BY-STEP EXPLANATION:

We need to factorise - x²y - x + 3xy + 3

Basic formula for factorisation is ax² + bx + c. In which we have to split the middle term in such a way that it's sum is 'b' and product is 'c'.

But in -x²y - x + 3xy + 3 we can't do this. So let's take common from it.

First put the xy terms on one bracket and rest in other bracket.

→ - x²y - x + 3xy + 3

→ (- x²y + 3xy) + (-x + 3)

Take xy common from x²y and 3xy & 1 from x and 3.

→ xy(-x + 3) +1(-x + 3)

→ xy(3 - x) +1(3 - x)

Take (3 - x) as common,

→ (3 - x) (xy + 1)

Therefore, the factors of -x²y - x + 3xy + 3 is (3 - x)(xy + 1).

ADDITIONAL INFORMATION:

Basic formulas for factorisation:

• a² - b² = (a - b)(a + b)
• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a³ - b³ = (a - b)(a² + ab + b²)
• a³ + b³ = (a + b)(a² - ab + b²)
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
• (a - b - c)2 = a2 + b2 + c2 - 2ab + 2bc - 2ca