Quadratic Factoring, Part 2

Last time I introduced the method to factor some quadratic equations so that you can use the Null Factor Law to solve the equation. The example used was

${x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}\hspace{0.33em}{=}\hspace{0.33em}{0}$

which when factored became

${(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}$

where you can readily see the two solutions x = 7 and -5. Let’s do another example.

${x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{24}\hspace{0.33em}{=}\hspace{0.33em}{0}$

So we know the factors will look like ${(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}$ and we need to find the a and b so that the factored form is equivalent to the left side of the equation. a and b must be numbers that multiply to equal 24 and add or subtract to equal +2, the coefficient in front of the x. 8 and 3 do not work as they do not add or subtract to equal 2. However, 6 and 4 look like contenders.

So the signs of 6 and 4 must be such that they add to equal +2 but multiply to equal -24. Looks like +6 and -4 work so

${x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}{24}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}$

where the two solution can be found by equating each factor to 0 which gives the two solutions x = 4 and -6.

Some quadratic equations may not be factorable but there is another method to solve these. I will show this method in my next post.

Now the generic quadratic equation looks like

${a}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{b}{x}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0}$

where ab, and c are specific numbers. I have been merciful so far in that I have only looked at equations where a = 1. If it is some other value, it’s a little more difficult to factor, but it can be done. However, I will use the quadratic formula to work with these in my next post.

Posted on Categories Algebra, Pre-VCE