A complex trinomial is of the form<\/p>\n
ax2<\/sup> + bx + c,<\/strong><\/p>\n where a\u22600,1. \u00a0An example is 2x2<\/sup> -3x -35. How do you factor such a trinomial?<\/p>\n Well, first of all, the easy trinomial method<\/a> won’t work for a complex trinomial. Let’s see, now, what will:<\/p>\n Example: Factor 2x2<\/sup> – 3x – 35<\/strong><\/p>\n Step 1:<\/strong> Multiply the lead coefficient (2 in this case) by the constant term (-35 in this case) to get -70.<\/p>\n Step 2:<\/strong> Find two numbers that multiply to make the product from step 1, but add to make the middle term coefficient (-3, in this case). Therefore, for our example, we need find the two numbers that multiply to make -70 but add to make -3. Of course, the numbers are -10 and 7.<\/p>\n Step 3:<\/strong> Rewrite the original trinomial, replacing the middle term with two terms whose coefficients are the numbers from step 2.<\/p>\n In other words, <\/p>\n 2x2<\/sup> -3x – 35 becomes<\/p>\n 2x2<\/sup> -10x +7x -35.<\/p>\n Step 4<\/strong><\/p>\n Common factor the first two terms from step 3. Then, common factor the last two. Do the pairs separately; it won’t be the same common factor for the first two as for the last two.<\/p>\n 2x(x-5) + 7(x-5)<\/p>\n Step 5<\/strong><\/p>\n Notice from Step 4 that, although the common factors you took out front don’t match, the brackets do match. Put the common factors in their own bracket, then rewrite:<\/p>\n (2x+7)(x-5)<\/p>\n Step 6<\/strong> (optional): Foil<\/a> out your answer from Step 5 to check it.<\/p>\n First: 2x(x)=2x2<\/sup><\/p>\n Outer: 2x(-5)=-10x<\/p>\n Inner: 7(x)=7x<\/p>\n Last: 7(-5)=-35<\/p>\n Add the four terms:<\/p>\n 2x2<\/sup> -10x +7x -35 = 2x2<\/sup> -3x -35<\/p>\n That’s one way to factor complex trinomials. Trial and error is faster; I’ll explore it in a future post:)<\/p>\n