Pertanyaan

A ka r − aka r p ers amaan 2 x 3 − 3 x 2 − 11 x + p = 0 a d a l ah x 1 , x 2 , d an x 3 . U n t u k x 1 = − 2 , ni l ai x 1 x 2 x 3 a d a l ah ...

$A ka r - aka r p ers amaan 2 x^{3} - 3 x^{2} - 11 x + p = 0 a d a l ah x_{1}, x_{2}, d an x_{3} . U n t u k x_{1} = - 2, ni l ai x_{1} x_{2} x_{3} a d a l ah ...$

-6
-3
0
3
6

D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

Menentukan nilai p terlebih dahulu, substitusikan x = - 2

Menentukan nilai p terlebih dahulu, substitusikan

x = - 2

$2 x cubed minus 3 x squared minus 11 x plus p equals 0 2 left parenthesis negative 2 right parenthesis cubed minus 3 left parenthesis negative 2 right parenthesis squared minus 11 left parenthesis negative 2 right parenthesis plus p equals 0 minus 16 minus 12 plus 22 plus p equals 0 p equals 28 minus 22 equals 6 S e h i n g g a colon 2 x cubed minus 3 x squared minus 11 x plus 6 equals 0 H a s i l space k a l i space k e t i g a space a k a r space u n t u k space b e n t u k space a x cubed plus b x squared plus c x plus d equals 0 space a d a l a h colon x subscript 1 x subscript 2 x subscript 3 equals fraction numerator negative D over denominator A end fraction equals fraction numerator negative 6 over denominator 2 end fraction equals negative 3$

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Teorema Sisa dan Faktor