Watch Video on Factoring Trinomials – Algebra Help
Thursday, November 19th, 2009
Access full lesson containing this video at: http://www.yourteacher.com/algebra1/factorpolynomials.php Students learn that a trinomial in the form x^2 + bx + c (where c is negative), such as x^2 + 6x — 27, can be factored as the product of two binomials, in this case (x + 9)(x — 3). The first term in each binomial comes from the factors of x^2, x and x. The second term in each binomial comes from the factors of the constant term, –27, that add to the coefficient of the middle term, +6, which in this case are +9 and –3.
Duration : 0:1:55
Access full lesson containing this video at: http://www.yourteacher.com/algebra1/graphinglinearinequalities.php Students learn to graph inequalities in two variables. For example, to graph y is less than x + 2, the first step is to graph the boundary line y = x + 2, using the chart method from lesson 4B. Note that greater than or less than means that the boundary line will be dotted, and greater than or equal to or less than or equal to means that the boundary line will be solid. To determine which side of the boundary line to shade, substitute a test point, such as (0, 0), into the original inequality, y is less than x + 2. Since (0) is less than (0) + 2, or 0 is less than 2, is a true statement, the side of the line that contains the point (0, 0) is shaded.
Access full lesson containing this video at: http://www.yourteacher.com/algebra1/completingthesquare.php Students learn to solve quadratic equations by completing the square. For example, to solve the equation s^2 – 6s + 5 = 0, the first step is put the constant term on the opposite side of the equation as the terms that contain the variables, by subtracting 5 from both sides, to get s^2 – 6s ___ = – 5 ___. Next, the number that goes in each space comes from half the coefficient of the middle term squared, which in this case is half of -6, or -3, squared, which is +9. So a +9 goes in each space, to get s^2 – 6s + 9 = – 5 + 9. The trinomial on the left side of the equation then factors as (s – 3)(s – 3), or (s – 3)^2, and the right side of the equation simplifies to 4, so the problem now reads (s – 3)^2 = 4. Finally, take the square root of both sides to get s – 3 = plus or minus 4, so s – 3 = 4, or s – 3 = -4, and solving each equation from here, s = 7 or -1.