Find the integral zeros of p(x) = x^4+x^3-5x^2+x-6.?

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2007-01-11 14:50:18 · 5 answers · asked by TM 3 in Science & Mathematics ➔ Mathematics

The integral roots are 2 and -3. You can easily check by synthetic division or by multiplying out (x - 2)(x + 3)(x^2 + 1)

2007-01-11 15:23:01 · answer #1 · answered by Pretzels 5 · 1⤊ 1⤋

x^4+x^3-5x^2+x-6 = 0

4x^3+3X^2-10x+1 = 0

12x^2 + 6x - 10 = 0

24x + 6 = 0

24x = -6

x= -6/24

x = -1/4

2007-01-11 14:56:11 · answer #2 · answered by Jo Jo 2 · 0⤊ 2⤋

1/5x^5+1/4x^4-5/3x^3+1/2x^2-6x+C

C is the unknown value that cannot be solved.

2007-01-11 14:53:46 · answer #3 · answered by Anonymous · 0⤊ 1⤋

I think the answer is 4

2007-01-11 14:53:05 · answer #4 · answered by Ted Arcidi 2 · 0⤊ 2⤋

Check JOJO's answer
i m sure he or she is rite.

2007-01-11 14:57:29 · answer #5 · answered by Anonymous · 0⤊ 2⤋