百度试题 结果1 题目8. Find the degree of polynomial.4+3x+6x^2-7x^3+8x^4 A)4 B)3 C)-7 D)-3 相关知识点: 试题来源: 解析 A 反馈 收藏
【题目】Find all rational zeros of the polynomial.(1)P(x)=x3+3x2-4(2) P(x)=x^3-3x-2(3) P(x)=x^3-6x^2+12x-8(4)P(x)=x3-4x2+x+6(5)P(x)=x3+3x2+6x+4(6)P(x)=x4-5x2+4(7) P(x)=x^4+6x^3+7x^2-6x-8(8) P(x)=4x^4-25x^2+36(9) P(x)=x^4+8x...
Find the zeros of the function algebraically. f(x) = 2x^2 - 3x - 20 Find the zeros of the function: f(x)=(8x+3)/(11-x) Find all the zeros of the function. f(x)=4x(x-3)^2 Determine the zeros of the function: 3x^2 - 8 = 10x. Find all zeros of the polynomial fu...
MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER All the real zeros of the given polynomial are integers. Find the zeros. (Enter your answers as a comma-separated list. Enter all answers including repetitions.) P(x) = x3 + 6x2 - 32 x = 2,- 2,...
P(x)=x4−6x3+7x2−6x+6; one zero = i. Zeros of Polynomials: A polynomial function will have as many zeros as the degree of the polynomial. Although these zeros might be real numbers, it is also possible that a quadratic factor will have no real zero...
The degree of a single term is the exponent of the variable if the term has only a single variable (or) the sum of the exponents of all the variables if it has multiple variables. The degree of a polynomial is the highest degre...
How do you factor the expression 12x^2 - 3? Multiply: (x^2 + x -1)(2x^2 - x + 2) Solve the polynomial. (7x^5-12x^2-6)-(5x^5+3x^2+9) How do you factor 12x^2 - 8x +1? What are the factors of 4x^4 + 8x^3 + 12x^2? Find the ...
How to determine if a polynomial has real roots? How to make a polynomial function with irrational zeroes? Find all rational zeros of the polynomial, and then find the irrational zeros, if any. P(x) = 2x^{4} + 13x^{3} + 20x^{2} + 11x + 2 How do you find the roots of a ...
Find the stationary points of the function {eq}f(x, y) = x^3 + xy^2 - 12x - y^2 {/eq} and identify their natures. Stationary Points The stationary points of a function are those where its derivatives vanish. For single variate functions, the...