3 Ways To Factor Algebraic Equations

Another way to think of this is that a given number’s factors are the numbers by which it is evenly divisible. Can you find all the factors of the number 60? We use the number 60 for a wide variety of purposes (minutes in an hour, seconds in a minute, etc. ) because it’s evenly divisible by a fairly wide range of numbers. The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

For example, the variable 12x can be written as a product of the factors of 12 and x. We can write 12x as 3(4x), 2(6x), etc. , using whichever factors of 12 are best for our purposes. We can even go as far as to factor 12x multiple times. In other words, we don’t have to stop with 3(4x) or 2(6x) - we can factor 4x and 6x to give 3(2(2x) and 2(3(2x), respectively. Obviously, these two expressions are equal.

Let’s try an example problem. To factor the algebraic equation 12 x + 6, first, let’s try to find the greatest common factor of 12x and 6. 6 is the biggest number that divides evenly into both 12x and 6, so we can simplify the equation to 6(2x + 1). This process also applies to equations with negatives and fractions. x/2 + 4, for instance, can be simplified to 1/2(x + 8), and -7x + -21 can be factored to -7(x + 3).

For example, let’s consider the algebraic equation. 5x2 + 7x - 9 = 4x2 + x - 18 can be simplified to x2 + 6x + 9 = 0, which is in the quadratic form. Equations with greater powers of x, like x3, x4, etc. can’t be quadratic equations. They are cubic equations, quartic equations, and so on, unless the equation can be simplified to eliminate these terms of x above the power of 2.

For example, let’s consider the quadratic equation x2 + 5x + 6 = 0. 3 and 2 multiply together to make 6 and also add up to make 5, so we can simplify this equation to (x + 3)(x + 2). Slight variations on this basic shortcut exist for slight variations in the equation itself: If the quadratic equation is in the form x2-bx+c, your answer is in this form: (x - _)(x - _). If it is in the form x2+bx+c, your answer looks like this: (x + _)(x + _). If it is in the form x2-bx-c, you answer is in the form (x + _)(x - _). Note: the numbers in the blanks can be fractions or decimals. For example, the equation x2 + (21/2)x + 5 = 0 factors to (x + 10)(x + 1/2).

Let’s consider an example problem. 3x2 - 8x + 4 at first seems intimidating. However, once we realize that 3 only has two factors (3 and 1), it becomes easier, because we know that our answer must be in the form (3x +/- _)(x +/- _). In this case, adding a -2 to both blank spaces gives the correct answer. -2 × 3x = -6x and -2 × x = -2x. -6x and -2x add to -8x. -2 × -2 = 4, so we can see that the factored terms in parentheses multiply to become the original equation.

For example, the equation x2 + 6x + 9 fits this form. 32 is 9 and 3 × 2 is 6. So, we know that the factored form of this equation is (x + 3)(x + 3), or (x + 3)2.

Let’s return to the equation x2 + 5x + 6 = 0. This equation factored to (x + 3)(x + 2) = 0. If either of the factors equals 0, the entire equation equals 0, so our possible answers for x are the numbers that make (x + 3) and (x + 2) equal 0. These numbers are -3 and -2, respectively.

Let’s plug -2 and -3 into x2 + 5x + 6 = 0. First, -2: (-2)2 + 5(-2) + 6 = 0 4 + -10 + 6 = 0 0 = 0. This is correct, so -2 is a valid answer. Now, let’s try -3: (-3)2 + 5(-3) + 6 = 0 9 + -15 + 6 = 0 0 = 0. This is also correct, so -3 is also a valid answer.

For example, the equation 9x2 - 4y2 = (3x + 2y)(3x - 2y).

The equation 4x2 + 8xy + 4y2 can be re-written as 4x2 + (2 × 2 × 2)xy + 4y2. We can now see that it’s in the correct form, so we can say with confidence that our equation factors to (2x + 2y)2

For instance, 8x3 - 27y3 factors to (2x - 3y)(4x2 + ((2x)(3y)) + 9y2)