Scan the code to test yourself and get the correct answers on Almanahj.
كويز تفاعلي: 6 Questions: قاعدة ديكارت والأصفار الحقيقية والتخيلية (ريفيل)
This assessment covers factoring polynomials and using Descartes' Rule of Signs to find possible zeros of functions. Key topics include synthetic division, factoring cubic expressions, and identifying positive, negative, and imaginary zeros.
🏆
Join the challenge and claim your rank
Take a timed certificate practice test for the same grade, subject, and term.
يرجى الانتباه إلى أن المعلم قام بإعداد الأسئلة فقط، ولم يقم بإعداد الإجابات أو الشروحات المرفقة. وقد تم توليد الإجابات باستخدام تقنيات الذكاء الاصطناعي، لذلك قد تتضمن بعض الأخطاء أو عدم الدقة.
للحصول على الإجابات الصحيحة والمضمونة، يُرجى الرجوع إلى المعلم أو المصدر الدراسي المعتمد.
Question 1
DB question no.: 1
Points: 1
Given one factor, find the remaining factors. x3 + 2x2 - x - 2; factor (x + 1). The other factors are?
Explanation
Dividing the polynomial x3 + 2x2 - x - 2 by (x + 1) results in x2 + x - 2, which factors into (x - 1)(x + 2).
Question 2
DB question no.: 2
Points: 1
x3 - 6x2 + 32; factor (x - 4). What are the remaining factors?
Explanation
Using synthetic division to divide x3 - 6x2 + 32 by (x - 4) gives x2 - 2x - 8, which factors as (x - 4)(x + 2). Since (x - 4) appears again, it is a double root.
Question 3
DB question no.: 3
Points: 1
State the possible number of positive real, negative real, and imaginary zeros. g(x) = 3x3 - 4x2 - 17x + 6: possible positive real zeros?
Explanation
According to Descartes' Rule of Signs, the number of positive real zeros is equal to the number of sign changes in g(x) or less by an even number. There are 2 sign changes ($+ \to -$ and $- \to +$), so there are 2 or 0 positive real zeros.
Question 4
DB question no.: 4
Points: 1
g(x) = 3x3 - 4x2 - 17x + 6: possible negative real zeros?
Explanation
Descartes' Rule of Signs for negative real zeros uses g(-x) = -3x3 - 4x2 + 17x + 6. In this polynomial, there is only 1 sign change ($-\to+$), meaning there is exactly 1 negative real zero.
Question 5
DB question no.: 5
Points: 1
h(x) = 4x3 - 12x2 - x + 3: possible positive real zeros?
Explanation
In h(x) = 4x3 - 12x2 - x + 3, there are 2 sign changes ($+\to-$ and $-\to+$), indicating either 2 or 0 positive real zeros.
For q(x) = x4 + 7x2 + 3x - 9, there is 1 positive real zero (1 sign change). For q(-x) = x4 + 7x2 - 3x - 9, there is 1 negative real zero (1 sign change). Since the polynomial has degree 4, it must have 4 total zeros. With 2 real zeros, there must be 2 imaginary zeros.
Here are more quizzes for الصف العاشر المتقدم by الفصل الثالث and subject رياضيات
This section is rendered only when the user reaches it while scrolling.
...
🍪
إشعار ملفات تعريف الارتباط
يستخدم هذا الموقع ملفات تعريف الارتباط لتحسين تجربة التصفح وقياس الأداء وعرض المحتوى بشكل أفضل.
باستخدامك للموقع فإنك توافق على استخدامنا لها وفق
سياسة الخصوصية.
