Solve By Factoring X^4-41x^2=-400

what are the real and complex solutions of the polynomial equation? x^3-8=0. 2, -1+i square root 3, and -1-i square root 3 find the roots of the polynomial equation. x^3-2x^2+10x+136=0. 3 +or- 5i, -4. which correctly describes the roots of the following cubic expressions? x^3-5x^2+3x+9=0. three real roots, two of which are equal in valueAnswer: The real solutions are -5, -4, 4, 5. There are no complex solutions. Step-by-step explanation: The equation.. x^4 -41x^2 +400 = 0. can be factored asFactoring Polynomial: Factoring of a polynomial is the process of reducing the polynomial to terms such that their multiplication gives the polynomial and the terms cannot be reduced any further.Complex Roots. If a polynomial has real coefficients, then either all roots are real or there are an even number of non-real complex roots, in conjugate pairs. For example, if 5+2i is a zero of a polynomial with real coefficients, then 5−2i must also be a zero of that polynomial.4.1 Solve x 4-41x 2 +400 = 0 This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using w , such that w = x 2 transforms the equation into : w 2-41w+400 = 0 Solving this new equation using the quadratic formula we get two real solutions : 25.0000 or 16.0000

what are the real and complex solutions of the polynomial

how many real number solutions does the equation have? 0=2x^2-20x+50 A. 1 solution B. 2 solutions C. no solutions** D. infinite solutions . Algebra . What are the real and complex solutions of the polynomial equation? x^4 - 41x^2= -400 . algerbaFree polynomial equation calculator - Solve polynomials equations step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy.II) Once a solution is found, Perform polynomial division until the the solution has been factored out. III) repeat I) and II) until the polynomial is of degree 2 and then try and either factorize the expression by inspection or by the quadratic formula. In the following polynomial this strategy doesn't get me very far.Complex Solutions to Polynomial Equations Starter 1. (Review of last lesson) Express the polynomials as the products of a linear and a quadratic factor with real coefﬁcients. Hence solve the equation when it is equal to zero. 2. In this question you must show detailed reasoning. (i.e. do not use your calculator)

$what are the real and complex solutions of the polynomial$

What are the real and complex solutions of the polynomial

We are given the polynomial equation x^4 - 41x^2 + + 1400 and is asked to determine the roots of the equation whether real or complex. The standard form should be the equal to the equation where -400 is up in the left side. The answers are 4 integers: 2 are negativeFind All Complex Solutions x^3+2x^2+4x+8=0. Factor the left side of the equation. Factor the polynomial by factoring out the greatest common factor, . If any individual factor on the left side of the equation is equal to , the entire expression will be equal to .It follows that all polynomial equations of degree 1 or more with real coefficients have a complex solution. On the other hand, an equation such as + = does not have a solution in (the solutions are the imaginary units i and -i). While the real solutions of real equations are intuitive (they are the x-coordinates of the points where the curveCalculator Use. This online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax 2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula. The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots.is called a polynomial equation of degree n. In this unit we are concerned with the number of solutions of polynomial equations, the nature of these solutions (be they real or complex, rational or irrational), and techniques for ﬁnding the solutions. We call values of x that satisfy equation (2) roots or solutions of the equation.

algebra

How many real number solutions does this equation have? 0= 5x^2 +2x - 12 one, two none, or infinite How many real quantity solutions does this equation have? 0= 2x^3 - 20x + 50

Algebra (need assist fast)

17. remedy the equation using the zero-product belongings. 6x(3x+1)=Zero A.x=0,1/3 B.0,1 C.0,-1/Three D.0,-1 18.Use the quadratic components to unravel the equation. If essential, spherical to the nearest hundredth. 6x(3x+1)=0 A.x=0,1/3 B.x=0,1

algebra

For the following equation, state the price of the discriminant and then describe the nature of the solutions. -2x^2+3x-7=0 What is the worth of the discriminant? Which one of the statements beneath is correct? The equation has two

Math

How many real quantity solutions does the equation have? y=3x^2-5x-Five A)one solution B)two solutions C)no solutions D)infinitely many solutions I can't figure this query out. Any lend a hand can be wonderful.