Question

plzz ans both the parts.
its urgent.

Answer 1

(2) 3Ycube-Ysquare-3Y+1
Ysquare(3Y-1)-1(3Y-1)
(3Y-1) (Ysquare-1square)
(3Y-1) (Y-1) (Y+1)

Answer 2

Hello Friend,

Here's the solution:

1) x³ + 13x² + 32x + 20

Here, sum of coefficients of terms with odd powers of x = 1+32 = 33

And, sum of coefficients of terms with even powers of x = 13+20 = 33

Both are same. This means that (x+1) is a factor.

So:
x³ + 13x² + 32x + 20
= x³ + x² + 12x² + 12x + 20x + 20
= x²(x+1) + 12x(x+1) + 20(x+1)
= (x + 1)(x² + 12x + 20)
= (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)

2) 3y³ - y² - 3y + 1

Here, sum of coefficients = 3-1-3+1 = 0
This means that (y-1) is a factor.

So,
3y³ - y² - 3y + 1
= 3y³ - 3y² + 2y² - 2y - y + 1
= 3y²(y-1) + 2y(y-1) -1(y-1)
= (y - 1)(3y² + 2y - 1)
= (y - 1)(3y² + 3y - y - 1)
= (y - 1)(3y(y+1) -1(y+1))
= (y - 1)(y + 1)(3y - 1)

Hope it helps.

Purva
@Purvaparmar1405
Brainly.in