百度试题 结果1 题目find the roots of the polynomial g(x)=x4+x3-8x2-2x+12 相关知识点: 试题来源: 解析 x^4+x^3-8x^2-2x+12 2 3 、 3 —2-6 0 0 6 0 —2 0 x^2-2 ∴g(x)=(x+3)(x-2)(x^2-2) 3,25,- 反馈 收藏
Find roots of polynomial equationsBill Venables
BThis can be done by factorizing the polynomial to (-1)(-2)(-3), so the roots are 1.2and 3which sum to 6.Another approach is to realize that (-)(-)(-)multiplies out to-(++)+(++)-, and as we're looking for++we can simply read it off the coefficient ofand flip it's ...
To find the multiple roots of the polynomial equation 8x3+20x2+6x−9=0, we will follow these steps: Step 1: Define the functionLet f(x)=8x3+20x2+6x−9. Step 2: Find the derivativeWe need to find the first derivative f′(x):f′(x)=ddx(8x3+20x2+6x−9)=24x2+40x+6...
Find all roots of the polynomial equation. {eq}\displaystyle x^5 + 3 x^4 - 2 x^3 - 14 x^2 - 15 x - 5 = 0 {/eq}. Roots of a polynomials: The roots of a polynomial are the numbers that make the polynomial expression equal to zero. If the polyn...
Find the rational roots of the polynomial 2x^3+3x^2-11 x-6 06:18 Show that x = 2 is a root of a polynomial x^3-6 x^2+11x-6 02:30 Find integral roots of the polynomial x^3 - 6x^2 +11x -6 04:48 If one zero of the polynomial p(x) = x^(3) - 6x^(2) + 11x - ...
{eq}\displaystyle f (x) = x^3 + 2 x^2 + 36 x + 72;\ (x + 2) {/eq} Remaining Roots of Polynomial Function: If the degree of a polynomial function is given and one of its factors is also given, then we can easily find the remaining r...
结果一 题目 100000÷100000000000000000 g(x)=x^4+x^3-8x^2-2x+12 答案 x^4+x^5-8x^2-2x+12 ##品3-2-6010—20x^2-2 ∴g(x)=(x+3)(x-2)(x^2-2) ∴ROdotaee^(-3,2),√2,-√2相关推荐 1100000÷100000000000000000 g(x)=x^4+x^3-8x^2-2x+12 ...
Consider the cubic polynomial p(x) = x3 - 6x2 + 11x - 6. Using NumPy, we can perform the following to determine this polynomial's roots ? Open Compiler import numpy as np p = [1, -6, 11, -6] roots = np.roots(p) print(roots) Output [3. 2. 1.] In this illustration, ...
Answer to: Graph this polynomial to find the roots x 3 + x 2 y + x y 2 + y 3 = 0 . By signing up, you'll get thousands of step-by-step solutions...