(x + 4)(x + 13)
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∙ 8y ago(x+13)(x+4)
Assuming this expression equals zero, it factorises as (x + 4)(x + 13) making x either -4 or -13.
x3 + x2 - 17x + 15 = (x - 1)(x - 3)(x + 5). Thus, the zeros are 1, 3, and -5. All three zeros are rational.
It is x+9, provided that x is not -8.
x + 4
(x+13)(x+4)
(x + 1)(x + 3)(x - 5)
(x + 3)(4x + 5)
Assuming this expression equals zero, it factorises as (x + 4)(x + 13) making x either -4 or -13.
4(x2 - 6x + 13)
It is x+9, provided that x is not -8.
x3 + x2 - 17x + 15 = (x - 1)(x - 3)(x + 5). Thus, the zeros are 1, 3, and -5. All three zeros are rational.
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
x + 4
x2(x3 + 1) is the best you can do there.
You can't factor it
That does not factor neatly.