\(3x^{2} + 2x - 4\)
\(x^{1} - 6x^{2} + x - 0\)
\(7x^{3} + x^{2} + x + 1\)
The key to adding polynomials is identifying the various like bases and then proceeding to add their coefficients. For instance, let’s add the following polynomials: \((x^{2} + 2x - 1) + (3x - 2x^{4} + 4)\)
Drop the parentheses, making the expression: \(x^{2} + 2x - 1 + 3x - 2x + 4\)
Next, identify the like terms. In this particular case, the like terms include \(x^{2}\) , \(x\), and the integer term.
Add the coefficient of each like base together to create the new term: \(-x^{2} + 5x + 3\)
When it comes to multiplying polynomials, the most important thing to remember is that each term from one polynomial will need to be multiplied with each term from the other polynomials.
Such an example of polynomial multiplication is \((Y + 3 - Y^{2})(2Y - 1 + 2Y^{2})\)
Y x 2Y = \(2Y^{2}\)
Y x -1 = -Y
Y x \(2Y^{2} = 2Y^{3}\)
3 x 2Y = 6Y
3 x -1 = -3
3 x \(2Y^{2} = 6Y^{2}\)
\(-Y^{2}\) x 2Y = \(-2Y^{3}\)
\(-Y^{2}\) x -1 = \(Y^{2}\)
-Y x \(2Y{2}\) = \(-2Y^{4}\)
\(2Y^{2} - Y + 2Y^{3} + 6Y - 3 + 6Y^{2} - 2Y^{3} + Y^{2} - 2Y^{4}\)
\(-2Y^{4} + 9Y^{2} + 5Y - Y - 3\)
When dividing polynomials, it is important to identify like bases and use the rule of exponents to simplify the terms which contain variables. Similarly to multiplying polynomials, when dividing polynomials each term in the dividend must be divided by each term in the divisor.
\((12x^{2} + 2x +4)\) ÷ (2x + 1)
Dividend: \(12x^{2}\) + 2x + 4
Divisor: 2x + 1
Performing this division will result in finding the quotient: \(2x + 1\sqrt{12x^{2} + 2x + 4}\)
In order to start dividing, look at the first term in the divisor and the first term in the dividend. In this particular case, 2x goes into \(12x^{2}\) , 6 times. In other words 2x x 6x = \(12x^{2}\) . Now multiply 6x to each term in the divisor.
From here, subtract the multiplied-out terms from the terms in the dividend.
Now, look at the second term in the dividend and see how many times the first term in the divisor can go into it. In this particular case, 2x goes into 2x one time. Thus we have:
From here, multiply the second terms of the quotient with the divisor and subtract from the dividend.
Since there are no terms left to subtract from, we place the remainder (-6x + 3) over the divisor in the quotient to create the final polynomial: \(6x + 1 + \frac{-6x + 3}{2x + 1}\)
Linear equations are equations in which all variables have an exponent of 1 and whose graph is a line.
A polynomial is an algebraic expression comprised of more than two terms, usually of like bases and different powers. Polynomials frequently include integer terms as well.
There are a few additional algebra topics often tested on the SAT. Expect each of the following to show up at least once on each SAT Math Test.
The Absolute Value of a number is its distance from zero (on the number line). Thus, an absolute value is always positive. We indicate absolute value by putting two bars around the number.
Students struggle with function based questions on the ACT. This chapter will give you all the information and solved examples you need to get the function questions correct
The term quadratic comes from the Latin word quadratus meaning square because the variable gets squared (e.g., \(x^{2}\) ). It is also called an equation of degree 2 because of the 2 superscripts on the x.
A TRANSFORMATION is a change of position or size of a figure. When we transform a figure, we create a new figure that is related to the original.