Find the zereos of following polynomial without the graph
1) x²-5x+6
2) x³-9x
Answers
Answered by
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1) x^2 - 5x +6 = 0
=> x^2 - 3x - 2x+6 = 0
=> x(x-3) - 2(x-3) = 0
=> (x-3)(x-2) =0
x= 2 and 3
2) x^3 - 9x = 0
=> x(x^2 - 9) = 0
=> x^2 - 9 = 0
=> x^2 = 9
x= 3 and - 3
=> x^2 - 3x - 2x+6 = 0
=> x(x-3) - 2(x-3) = 0
=> (x-3)(x-2) =0
x= 2 and 3
2) x^3 - 9x = 0
=> x(x^2 - 9) = 0
=> x^2 - 9 = 0
=> x^2 = 9
x= 3 and - 3
Answered by
0
hey here is your answer
1) by splitting middle term
x²-5x+6
x²-3x-2x+6
x(x-3)-2(x-3)
(x-2)(x-3)
x = 2 & x = 3
2) x³-9x=0
x³ = 9x
x² = 9
x = 3
hope it helps u ☺
1) by splitting middle term
x²-5x+6
x²-3x-2x+6
x(x-3)-2(x-3)
(x-2)(x-3)
x = 2 & x = 3
2) x³-9x=0
x³ = 9x
x² = 9
x = 3
hope it helps u ☺
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