Search results
Results From The WOW.Com Content Network
In an example application of this procedure, (3x + 5)/(1 − 2x) 2 can be decomposed in the form + = + (). Clearing denominators shows that 3x + 5 = A + B(1 − 2x). Expanding and equating the coefficients of powers of x gives
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
In polynomials with one indeterminate, the terms are usually ordered according to degree, either in "descending powers of x", with the term of largest degree first, or in "ascending powers of x". The polynomial 3x 2 − 5x + 4 is written in descending powers of x. The first term has coefficient 3, indeterminate x, and exponent 2.
The graph of the logarithm to base 2 crosses the x axis (horizontal axis) at 1 and passes through the points with coordinates (2, 1), (4, 2), and (8, 3). For example, log 2 (8) = 3, because 2 3 = 8. The graph gets arbitrarily close to the y axis, but does not meet or intersect it.
The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.According to the theorem, the power (+) expands into a polynomial with terms of the form , where the exponents and are nonnegative integers satisfying + = and the coefficient of each term is a specific positive integer ...
Divide the first term of the dividend by the highest term of the divisor (x 3 ÷ x = x 2). Place the result below the bar. x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by ...