![]() ![]() If you can factor this into the product of even more expressions. To factor six x squared plus nine x times x squared ![]() What we're going to do in this video is do a few more examples of factoring higher degree polynomials. ![]() Many videos on Khan Academy where we talk about factoring polynomials. Additionally, for some higher-degree polynomials, finding exact solutions may not be feasible or practical, and numerical methods are the best approach. For these higher degrees, numerical methods are often used to approximate the roots. It's essential to note that there is no general algebraic formula (analogous to the quadratic formula for degree 2 polynomials) for finding roots of polynomials with degrees higher than 4. * Numerical Methods*: Besides root-finding algorithms, there are other numerical methods like polynomial interpolation and numerical integration techniques that can help approximate the roots of higher-degree polynomials. * Graphical Methods*: For a rough estimation of the roots, you can plot the graph of the polynomial and identify the approximate values where it crosses the x-axis, representing the roots.Ħ. It doesn't directly give the roots but narrows down the possibilities.ĥ. * Descartes' Rule of Signs*: This rule provides information about the possible number of positive and negative real roots of a polynomial by counting the sign changes in its coefficients. By completing the square, you can transform certain higher-degree polynomials into a quadratic form that can then be solved using the quadratic formula.Ĥ. * Completing the Square*: This method is commonly used for solving quadratic equations (polynomials of degree 2) but can be extended to higher-degree polynomials in some cases. These methods iteratively refine an initial guess until an accurate root is found.ģ. * Root-Finding Algorithms*: For polynomials that cannot be easily factored, numerical root-finding algorithms like the Newton-Raphson method, bisection method, or the secant method can be used to approximate the roots. However, it becomes more challenging for higher-degree polynomials with no simple factorization.Ģ. Factoring works well for polynomials with rational roots or when they can be factored into binomial or trinomial expressions. * Factoring*: This method involves factoring the polynomial into simpler expressions that can be set to zero to find the roots (solutions). Yes, there are several methods to solve higher-degree polynomials (polynomials of degree three or higher) other than grouping. Also, it means you just did all of that math to get a circle (start in one place, end in the same). That means that this is as simplified as you can get your equation. When you multiply that out, you get x^3 - 4x^2 + 6x + 24. This means that we can take the numbers outside the parentheses and put them in their own set. As you can see, the sets of numbers inside the parentheses are the same. Now we take out the GCF from both equations and move it to the outside of the parentheses. To do that, you put parentheses around the first two terms and the second two terms. The easiest way to solve this is to factor by grouping. The second example is a little different: This gets usģx(2x + 3)(x - 2)(x - 2) Since you can no longer factor this equation, it is in simplest form. The next step is to put all of that together. Splitting (x^2 - 4x + 4) into its square roots results in this: The other we can tell just by looking that it is a perfect square, so we split it apart as shown in the first unit called Polynomial Arithmetic, with the video Polynomial special products: perfect square. Now we bring the 3x to the outside of the parentheses to get 3x(2x + 3). This is better illustrated in the video Taking common factor from binomial, under Taking common factors earlier in this unit, so I am going to skip explaining that part. Now that we have our GCF for the left set of parentheses, we can divide everything in the left set by our GCF, and bring the GCF out of the parentheses. Since the GCF of the variables is x and the GCF of the coefficients is 3, we multiply them together to get 3x. We can see by looking at the factors that the greatest factor is x (assuming x is greater than one, but we aren't going to worry about that because we don't have to solve for x in this problem). X^2 factors are 1 and x^2, and x and x since those numbers multiply to get to x^2. The next step to find the GCF of the full terms is to look at the variables. As you may have seen in previous videos in this unit, the way to find the GCF from the left set of parentheses is to find the GCF of the coefficients. What Sal did was take the GCF out of each set of parentheses.
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