Question

x ^2 - 19 X + 88 find zeros of polynomial​

Answer 1

Step-by-step explanation:

x²-19x+88=0

x²-11x-8x+88=0

x(x-11)-8(x-11)

(x-8)(x-11)

x-8=0

x=8

x-11=0

x=11

zeroes of the polynomial x²-19x+88 are 8 and 11.

Answer 2

Answer:

hi,

Step-by-step explanation:

x²-19x+88

x²-(11+8)x+88

x²-11x-8x+88

x(x-11)-8(x-11)

(x-8)(x-11)