Question

please answer quickly ​

Answer 1

⏩3x²-4x-4

☞3x²-6x+2x-4

☞3x(1x-2) 2(1x-2)

$\boxed{\pink{(1x-2) \: (3x-2)}}$

⏩30m²+m-3

☞30m²+10m-9m-3

☞10m(3m+1) -3(3m+1)

$\boxed{ \pink{(3 m+1) \: (10m-3)}}$

⏩-14p²-5p+1

☞-14p²-7p+2p+1

☞-7p(2p+1) +1(2p+1)

$\boxed{ \pink{ (2p + 1) \: (-7p + 1)}}$

⏩6x²+11x+5

☞6x²+6x+5x+5

☞6x(x+1) 5(x+1)

$\boxed{ \pink{(x+1) \: (6x + 5)}}$

⏩23x²-25x+2

☞23x²-23x-2x+2

☞23x(x-1) -2(x-1)

$\boxed { \pink{(x - 1) \: (23x - 2)}}$

⏩4p³-2p²+2p-1

☞2p²(2p-1) +1(2p-1)

$\boxed{ \pink {(2p - 1) \: ({2p}^{2} + 1)}}$

⏩m³+m²-4m-4

☞m²(m+1) -4(m+1)

☞(m²-4) (m+1)

☞[(m)²-(2)²] (m+1)

$\boxed{ \pink{(m + 2) \: (m - 2) \: (m + 1)}}$

⏩10p³-p²+100p-10

☞p²(10p-1) +10(10p-1)

$\boxed{ \pink{(10p - 1) \: ({p}^{2} + 10)}}$

⏩8y³+y²-8y-1

☞y²(8y+1) -1(8y+1)

$\boxed{ \pink{(8y + 1) \: ({y}^{2} - 1)}}$

⏩2x³-3x²-17x+30

☞2x³- 4x + x²- 2x- 15x+ 30

☞2x²(x-2) +x(x-2) -15(x-2)

☞(x-2) (2x²+x-15)

☞(x-2) (2x²+6x-5x-15)

☞(x-2) [2x(x+3) -5(x+3)]

$\boxed { \pink{(x - 2) \: (x + 3) \: (2x - 5)}}$

✝️$\orange{Some \: Rules:-}$

• Factorisation is the reverse of multiplication.

• Taking out the common factors:- When each term of the given equation contains a common factor divide each term by this factor and enclose the quotient within brackets, keeping the factor outside the bracket.

• By grouping:- An expression of an even number of terms may be resolved in two factors if the terms are arranged in groups such that each group has a common factor.

• By splitting the middle term:- when a trinomial is in the form ay²+bx+c, of split b (the coefficient of x in the middle term) into two parts such that the sum of these two parts is equal to b and the product of these two parts is equal to the product of a and c. Then factorise by the grouping method.