Factoring a trinomial with a lead coefficient, not one, resources.
In Algebra a trinomial is a polynomial with three terms. Use this method I call the "rainbow method" to help with your factoring.
This rainbow method works great with a trinomial with a led coefficient greater than one, and there is not a greatest common factor.
Example problem 1 Factor 6x^2 +7x -5
Example problem 2 Factor 15x^2 + 17x -4
Hi welcome to MooMooMath Today we are going to look at factoring a trinomial that has a lead coefficient that is not one. I call this the rainbow method. It will help if you know how to factor by grouping, but if you don't it is ok, but we use factor by grouping and it is helpful if you know it.
The first thing we need to do is identify which trinomials we can use the rainbow method with. If our lead coefficient is greater than one (in this example it is a 6) and there is not a greatest common factor, then it is a great time to use the rainbow method. To use the rainbow method I'm going to take a because this is in standard form Let,s write our standard form so you can see it. I will multiply a times c, and see how I wrote it over top. This is why it is the rainbow method. Take time c and you get negative 30. Next, ask yourself what multiplies to negative 30, middle term but adds to the middle term. Our middle term b is positive 7. So we are looking for two values that multiply to negative 30 but add to 7. So let's find some combination. 1x30 no,2x15,no 3 times 10. Yes. Here is a combination that multiplies together to equal 30 and adds to positive 7. So the signs will be 3 and 10 are the two numbers, and the sign is negative so the three must be negative to get the 30. Now what I will do is take the two factors, middle term, the 7x and split it into two parts. A negative 3x +10x I 'm rewriting the middle term to be two different terms which are the sum of the two required factors. I will then bring down my a squared which is 6x^2. I then will bring down my c which is negative 5. The front term and the back term remain the same. Now I will factor by grouping because I have four terms. Factor the first two terms together, and the last two terms together. When I do this I will take the greatest common factor out of the first two terms which are 3x , and I left with 2x minus 1. I will do the same to the back. What is my common factor? 5 I will factor with positive 5 and I'm left with (2x -1) Now what is nice about factoring by grouping is that you can see these two factors need to match and they do. I will now factor the (2x-1) out. I will take the (2x-1) out of both terms, and when I do I'm left with (3x +5.) And there is how I factor by grouping. Lets do one more of those. 15x^2 + 17x -4
Does it qualify, yes the first term is greater than 1, and there is not a greatest common factor? Multiply the first term by the last term. 15 times 4 = 60. Next, what multiplies together to equal 60, but adds together to equal the middle term. 3 times 20 equals 60, and when you subtract 20-3 they equal the middle term. The 20 has to be positive, and the 3 has to be negative. Let's split the middle number to be -3x plus 20x. The order I write these down doesn't matter. Bring down c and bring down an x squared. Now factor by grouping. Group the first two together. What is my common factor? 3x is my common factor. I'm left with (5x -1) What is common in the back? 4 is common and I'm left with (5x-1). Do my binomials match? Yes, Factor out this binomial (5x-1) and we are left with (5x-1)(3x+4) I hope this rainbow method has been helpful when your lead coefficient is not one.
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Tagged under: Trinomial,Algebra (Field Of Study),factoring,trinomials,factor grouping,factoring polynomials,algebra,Factorization (Literature Subject)
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