Can you solve this algebra problem

1) cube root of 12x^2 ( cube root of 126x^2)

2) Find the zeros of y = x^3 -11x^2+ 41x - 51

Please show your steps!

Cube root of 126x² is in its simplest form

Could it have been cube root of 128x²

because 128 = 64 x 2

......................................…
so cube root of 128x² = 4\³/(2x²)


Now to get zeros of x² -11x² +41x-51

we shall first note that 51 = 17x3

then determine if 1 or -1 is a zero

when x=1

y=1³-11(1)² +41(1)-51 = 1-11+41-51 = -10+41-51=-20

lets try x=-1 giving us

y=(-1)³-11(1)² +41(-1)-51 = -1-11-41-51 = -12-41-51 which is definitely a negative number

let us now try x=3

y=(3)³-11(3)² +41(3)-51 = 27 -99+ 123-51=66 =0

boom! That is a zero. x=3 is one of the zeros and

x-3 is a factor

divide x³ -11x² +41x-51 by x -3

......x²..-8x..+17
...._______________
x-3)x³ -11x² +41x-51
.....x³ - 3x²......↓..Subtract this bring down
....————...........the 41x
..........-8x² + 41x
..........-8x²..+24x Subtract this then bring down
........————...........the 51
...................17x - 51
...................17x - 51 subtract this
..................————
.........................zero remainder


so

x²..-8x..+17 is quotient

the other zeroes can be obtained by completing
the square or using quadratic formula

If we complete square on

x²..-8x..+17 = 0
we get

x²..-8x. . = -17
x² - 8x + 16 = -17 + 16 = -1

so

(x+ 4)² = -1

thus

x+4 = ±i

so
x = -4+i or x= -4-i

thus all zeros, including complexes are

x=3, x=-4+i , x= -4-i

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