Solving polynomial equations algebraically is a fundamental skill in algebra that involves various techniques to find the roots or zeros of a function. For quadratic equations, factoring or the quadratic formula are common approaches. For higher-degree polynomials, techniques such as factoring by grouping, using the sum and difference of cubes formulas, and applying the Rational Root Theorem become essential. The Fundamental Theorem of Algebra states that a polynomial of degree n will have exactly n complex roots, though some may be repeated. Understanding the relationship between the factors of a polynomial and its x-intercepts on a graph allows for a deeper comprehension of algebraic structures and their behaviors.
رقم الاختبار827
الصفالصف العاشر المتقدم
المادةرياضيات
الفصلالفصل الثالث
السنة الدراسية2025/2026
عدد الأسئلة28
إجمالي النقاط28
تاريخ الإضافة2026-04-21
الزيارات49
المعلم أو الناشرAmal Salman
اختر إجابة واحدة لكل سؤال. عند الاختيار ستظهر النتيجة فورًا: الأخضر صحيح، والأحمر خطأ، وسيظهر تفسير الإجابة مباشرة إن كان متوفرًا. وبعد آخر سؤال ستظهر الدرجة النهائية تلقائيًا.
Question 1
Points: 1
Solve by Factoring: \(4x^2 + 11x - 3 = 0\)
Explanation
Factoring the quadratic equation \(4x^2 + 11x - 3 = 0\) results in \((4x - 1)(x + 3) = 0\). Setting each factor to zero gives \(x = 1/4\) and \(x = -3\).
Question 2
Points: 1
Solve by Factoring: \(x^2 + 15x + 64 = 20\)
Explanation
Subtract 20 from both sides to get \(x^2 + 15x + 44 = 0\). Factoring gives \((x + 11)(x + 4) = 0\), so the solutions are \(-11\) and \(-4\).
Using the Remainder Theorem, evaluate \(P(-5) = (-5)^3 + 2(-5)^2 - (-5) + 4 = -125 + 50 + 5 + 4 = -66\). Since the remainder is not zero, \(x + 5\) is not a factor.
Question 22
Points: 1
Solve by factoring: \(3x^4 - 6x^2 + 3 = 0\) (Hint: take out the GCF)
Explanation
Factor out 3: \(3(x^4 - 2x^2 + 1) = 0\), which is \(3(x^2 - 1)^2 = 0\), and further \(3(x - 1)^2(x + 1)^2 = 0\). The roots are \(1\) and \(-1\), both as double roots.
Question 23
Points: 1
How many x-intercepts does \(x^4 + 6x^2 + 9 = 0\) have?
Explanation
The polynomial is \((x^2 + 3)^2 = 0\). The roots are \(\pm i\sqrt{3}\), which are imaginary. Therefore, there are no real x-intercepts. However, the answer key provided in the document indicates 4.
Question 24
Points: 1
Is \((x-3)\) a factor of \(x^3 - 9x^2 + 25x - 21\)?
Explanation
By the Factor Theorem, evaluate \(P(3) = 3^3 - 9(3)^2 + 25(3) - 21 = 27 - 81 + 75 - 21 = 0\). Since the result is zero, \(x - 3\) is a factor.
Question 25
Points: 1
How many roots does the polynomial function \(g(x) = x^4 - 3x^2 + 10\) have?
Explanation
According to the Fundamental Theorem of Algebra, a polynomial of degree \(n\) has exactly \(n\) roots (including real and complex roots). Since the degree is 4, it has 4 roots.
Question 26
Points: 1
How many solutions does the polynomial equation \(5x^3 - 10x^2 + 8x + 1 = 0\) have?
Explanation
The degree of the polynomial is 3, which means it has exactly 3 solutions in the complex number system.
Question 27
Points: 1
How many solutions does the polynomial equation \(3x^2 - 5x^4 + 3x - 7 = 0\) have?
Explanation
The degree of the polynomial (the highest power of \(x\)) is 4, which indicates there are 4 solutions.
Question 28
Points: 1
Which ONE of the following is NOT a possible solution to the polynomial equation \(3x^4 - 6x^3 + 2x + 4 = 0\)?
Explanation
Using the Rational Root Theorem, possible rational roots are of the form \(\pm p/q\) where \(p\) is a factor of the constant term (4) and \(q\) is a factor of the leading coefficient (3). Factors of 4 are {1, 2, 4} and factors of 3 are {1, 3}. \(-3\) is not among the possible roots.
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