Factoring Trinomials | Math Placement Test Prep | MathHelp.com @yourteachermathhelp

Uploaded May 2026 | Updated September 2026, 2 weeks ago

Factoring trinomials is part of a complete math placement test prep course at MathHelp.com:
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This lesson covers factoring trinomials. 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.

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$Midpoint Formula | MathHelp.com Need a custom math course? Visit https://www.MathHelp.com Today! In this example, we’re asked to find the midpoint of the line segment joining the points (10, 1) and (2, -7), so we use the midpoint formula, which states that the midpoint of the line segment joining two points is parentheses x1 + x2 over 2…comma… y1 + y2 over 2 closed parentheses. Our first point, (10, 1), represents (x1, y1), and our second point, (2, -7), represents (x2, y2). So plugging the given information into the formula, we have parentheses x1, or 10, + x2, or 2, over 2…comma…y1, or 1, + y2, or -7, over 2, closed parentheses. Next, simplifying in the numerators, 10 + 2 is 12, so we have 12/2, and 1 + -7 is -6, so we have -6 over 2. Finally we simplify our fractions. 12/2 is 6, and -6 over 2 is -3. So the midpoint of the line segment joining the points (10, 1) and (2, -7) is (6, -3).$

$Imaginary Numbers | MathHelp.com Need a custom math course? Visit http://www.MathHelp.com. This lesson covers imaginary numbers. Students learn that the imaginary number i is equal to the square root of -1, which means that i^2 is equal to (the square root of -1) squared, which equals -1. Students also learn to simplify imaginary numbers. For example, to simplify the square root of -81, think of it as the square root of -1 times the square root of 81, which simplifies to i times 9, or 9i. To simplify 11/8i, the first step is to get rid of the i in the denominator by multiplying both the numerator and the denominator of the fraction by i, to get 11i/8i^2, and remember that i^2 = -1, so we have 11i/8(-1), or 11i/-8, or -11i/8.$

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