Math, asked by dhavaljain06510, 11 months ago

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

Answers

Answered by krishnmnaik
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 pantnagarlucknow
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)

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