Math, asked by LakshyaS2006, 2 months ago

factorise by splitting the middle term: x³-2x²-x+2 x³-3x²-9x-5
x³-13x²+32x+20 2y³+y²-2y-1​

Answers

Answered by dakshkuyadav85
1

Answer:

x³-2x²-x+2→

x³-2x²-x+2

=> x²(x-2)-1(x-2)

=> (x²-1)(x-2)

(1 can be Expressed as 1²)

=> (x²-1²)(x-2)

(a²-b² = (a+b)(a-b))

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

The factors of x³-2x²-x+2 are→

(x+1)(x-1)(x-2)

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2.) x + 1 is the factor.

x^3 + x^2 - 4x^2 - 4x - 5x - 5

x^2 (x + 1) - 4x (x + 1) - 5 (x + 1)

(x + 1) (x^2 - 4x - 5)

(x+1)(x² - 5x + x - 5)

= (x + 1) (x - 5) (x + 1).

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3.) x³+13x²+32x+20

=(x+1)(x²+12x+20) [∵, for x=-1, x³+13x²+32x+20=-1+13-32+20=0]

=(x+1)(x²+10x+2x+20)

=(x+1){x(x+10)+2(x+10)}

=(x+1){(x+10)(x+2)}

=(x+1)(x+2)(x+10)

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4.) 2y³ + y² - 2y -1

= 2y^3 - 2y + y^2 - 1

= 2y(y^2 - 1) +1(y^2 - 1)

= (2y+1) ( y^2 - 1)

= (2y+1) ( y^2 - 1^2)

= (2y+1) ( y+1) ( y-1)

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