please solve this problem correctly ........
Answers
Answer:
Here is your answer bro
Step-by-step explanation:
A} x3-2x2-x+2
Ans: x³-2x²-x+2
x²(x-2)-1(x-2)
(x²-1)(x-2)
B} x3-3x2-9x-5
Ans: x³ - 3x² - 9x - 5
= x³ + x² - 4x² - 4x - 5x - 5
= x²( x + 1) - 4x ( x + 1) - 5(x + 1)
= (x + 1)(x² - 4x - 5)
= (x + 1)(x² -5x + x - 5)
= (x + 1)(x - 5) (x + 1)
Hence, (x +1), (x -5) and (x +1) are the factors of given polynomial.
C} x3+13x2+32x+20
Ans: 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)
D} 2y3=y2-2y-1
Ans: (2y+1)(y+1)(y-1)
=2y³+y²-2y-1
= y²(2y+1)-1(2y+1)
= (2y+1)(y²-1)
= (2y+1)(y²-1²)
= (2y+1)(y+1)(y-1)
By algebraic identity:
a²-b² = (a+b)(a-b) */
Therefore,
2y³+y²-2y-1=(2y+1)(y+1)(y-1)
Please mark me as the brainliest.