x ^2 - 19 X + 88 find zeros of polynomial
Answers
Answered by
5
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.
Answered by
4
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)
Similar questions