اختبار إلكتروني: Solving Polynomial Equations by Graphing and Factoring
This worksheet focuses on various methods for solving and factoring polynomial equations. Topics include factoring by grouping, using the greatest common factor (GCF), factoring differences of squares, and finding the zeros of functions from equations or graphs. It also covers identifying the number of solutions based on the degree of the polynomial and dealing with complex solutions.
رقم الاختبار826
الصفالصف العاشر المتقدم
المادةرياضيات
الفصلالفصل الثالث
السنة الدراسية2025/2026
عدد الأسئلة21
إجمالي النقاط21
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
To solve \(4x^2 + 11x - 3 = 0\) by factoring, find two numbers that multiply to \(-12\) and add to \(11\), which are \(12\) and \(-1\). This gives \((4x - 1)(x + 3) = 0\), resulting in \(x = \frac{1}{4}\) and \(x = -3\).
This is a difference of squares: \(a^2 - b^2 = (a - b)(a + b)\). Since \(400 = 20^2\), \(x^2 - 400 = (x - 20)(x + 20)\).
Question 4
Points: 1
Solve \((x^2 - 9) = 0\)
Explanation
\(x^2 - 9 = 0\) implies \(x^2 = 9\). Taking the square root of both sides gives \(x = 3\) and \(x = -3\).
Question 5
Points: 1
Solve by factoring: \(3x^4 - 6x^2 + 3 = 0\) (Hint: take out the GCF)
Explanation
Factoring out the GCF: \(3(x^4 - 2x^2 + 1) = 0\). This is a perfect square trinomial: \(3(x^2 - 1)^2 = 0\). Factoring further: \(3((x - 1)(x + 1))^2 = 0\), which gives roots \(x = 1\) and \(x = -1\), each with multiplicity 2.
Question 6
Points: 1
Solve: \(2x^2 - 14x + 24 = 0\)
Explanation
Divide by 2: \(x^2 - 7x + 12 = 0\). Factoring the quadratic: \((x - 4)(x - 3) = 0\), so \(x = 4\) or \(x = 3\).
Question 7
Points: 1
Find the solutions to the following: \(x^2 = -9\)
Explanation
Taking the square root of both sides: \(x = \pm \sqrt{-9} = \pm 3i\).
Question 8
Points: 1
What is the first step when factoring polynomials?
Explanation
The standard procedure for factoring any polynomial is to first check for the Greatest Common Factor (GCF) and factor it out if it exists.
Question 9
Points: 1
Factor Completely: \(8x^3 - 18x\)
Explanation
First factor out the GCF \(2x\) to get \(2x(4x^2 - 9)\). Then recognize that \(4x^2 - 9\) is a difference of squares, which factors into \((2x + 3)(2x - 3)\).
Question 10
Points: 1
How many solutions does the following expression have? \(4w^4 + 16w^2 + 15\)
Explanation
According to the Fundamental Theorem of Algebra, a polynomial of degree \(n\) has exactly \(n\) complex roots (counting multiplicity). Here the degree is 4.
Question 11
Points: 1
How many solutions does the following expression have? \(7x^5 - 3x^4 + 5x - 8\)
Explanation
The number of solutions is determined by the highest power of the variable (the degree of the polynomial). Since the degree is 5, there are 5 solutions.
Question 12
Points: 1
f(x) = (x+4)(x-3)(x-2) List the zeros for this function.
Explanation
Zeros occur where each factor equals zero. \(x + 4 = 0 \implies x = -4\); \(x - 3 = 0 \implies x = 3\); \(x - 2 = 0 \implies x = 2\).
Question 13
Points: 1
How many zeros does the polynomial function \(P(x) = 3x^4 + 4x - 8\) have?
Explanation
The degree of the polynomial \(P(x)\) is 4, which means the function has 4 zeros in the complex number system.
Find all zeros for the polynomial function \(P(x) = x^3 - 3x^2 - 5x + 15\)
Explanation
Factoring by grouping: \(x^2(x - 3) - 5(x - 3) = 0 \implies (x^2 - 5)(x - 3) = 0\). The zeros are \(x = 3\) and \(x = \pm \sqrt{5}\).
Question 16
Points: 1
Find all zeros for the polynomial function \(P(x) = x^3 + 6x^2 + 9x + 54\)
Explanation
Factoring by grouping: \(x^2(x + 6) + 9(x + 6) = 0 \implies (x^2 + 9)(x + 6) = 0\). The zeros are \(x = -6\) and \(x = \pm \sqrt{-9} = \pm 3i\).
Question 17
Points: 1
Solve the following equation: \(x^4 + x^2 - 90 = 0\)
Explanation
Let \(u = x^2\), then \(u^2 + u - 90 = 0 \implies (u + 10)(u - 9) = 0\). Thus \(x^2 = 9 \implies x = \pm 3\), and \(x^2 = -10 \implies x = \pm i\sqrt{10}\). Note: The provided answer key chooses option (a), though the square root of -10 is imaginary.
Question 18
Points: 1
Solve the following equation: \(x^4 - 6x^2 + 8 = 0\)
Explanation
Let \(u = x^2\), then \(u^2 - 6u + 8 = 0 \implies (u - 4)(u - 2) = 0\). Thus \(x^2 = 4 \implies x = \pm 2\), and \(x^2 = 2 \implies x = \pm \sqrt{2}\).
Question 19
Points: 1
What are the zeros of the function?
Explanation
The zeros of the function are the x-intercepts of the graph. Looking at the provided graph, the curve crosses the x-axis at \(x = -3\), \(x = -1\), \(x = 2\), and \(x = 5\).
Question 20
Points: 1
Solve by factoring: \(14x - 49x^2 = 0\)
Explanation
Factor out \(7x\): \(7x(2 - 7x) = 0\). This gives \(7x = 0 \implies x = 0\), and \(2 - 7x = 0 \implies x = \frac{2}{7}\).
Question 21
Points: 1
Solve by factoring: \(3x^3 + x^2 - 14x = 0\)
Explanation
Factor out \(x\): \(x(3x^2 + x - 14) = 0\). Factoring the quadratic part gives \(x(3x + 7)(x - 2) = 0\). The roots are \(x = 0\), \(x = -\frac{7}{3}\), and \(x = 2\).
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