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Well, the easiest way to solve this is to use a graphing calculator. However, for the time being, let's assume we don't have that resource available to us.
So, let's start by listing possible rational roots. Now, by the rational root theorem (see link below), our roots will be of the form:
`(x,0)| x = +- p/q`
Where `p` is an integer factor of the constant term, and `q` is an integer factor of the x term of highest degree. Therefore, we get the following possibilities for p and q:
`p = 1, 2, 4, 8`
`q = 1`
So, our possible rational roots will be:
`x = +- 1, +-2, +-4, +-8`
Now, we can actually eliminate these by using Decartes' Rule of Signs.
Notice that there are no sign changes in the equation as it is, indicating there are zero positive roots! This reduces our possibilities by half!
`x = -1, -2, -4, -8`
So, let's go down the line. Let's see if -1 is a root by dividing the function by (x+1) and seeing if we have a remainder:
`(x^3 + 7x^2 + 14x +8)/(x+1) = x^2+6x+8`
Looks like it works out! I actually did this by long division, which isn't easily demonstrated here, but I encourage you to try it on your own! It's a valuable skill in Algebra II!
So, we have a partially factored function:
` ``f(x) = (x+1)(x^2+6x+8)`
Factoring the trinomial is much easier. We just need to find two numbers that add to +6 and that multiply to +8. This sounds like 2 and 4 to me! Let's show our completely factored form:
`f(x) = (x+1)(x+2)(x+4)`
Now, remember, our roots are going to be the negative forms of the constants inside the parantheses! This means we get the following solutions to the equation:
`x = -1, -2, -4`
Let's confirm by graphing:
Looks like we're correct!
I hope that helps!
You should use the factorization method when notice that groupping terms yields common factors.
In this case, the middle terms may be collected, also the first and the last terms, hence you will get two groups such that:
`(x^3 + 8) + (7x^2 + 14x) = 0`
Notice that 8 is the third power of 2, hence `x^3 + 8 = x^3 + 2^3` .
You should substitute the sum of cubes by the product `(x+2)(x^2-2x+4)` , hence:
(`x+2)(x^2-2x+4) + (7x^2 + 14x) = 0`
Notice that the terms in the group (`7x^2 + 14x` ) share the common factor 7x such that:
`(x+2)(x^2-2x+4) + 7x(x + 2) = 0`
You may factor out `x+2` such that:
`(x+2)(x^2-2x+4+7x) = 0`
`` `x+2 = 0 =gt x_1 = -2`
`` `x^2 + 5x + 4 = ` 0
You should write `5x = 4x + x` , hence:
`x^2 + x + 4x + 4 = 0 =gt x(x+1) + 4(x+1) = 0`
`` `(x+1)(x+4) = 0`
`` `x+1 = 0 =gt x_2 = -1`
`` `x + 4 = 0 =gt x_3 = -4`
Sketching the graph of the function you should notice that the curve intercepts x axis 3 times, exactly at the values -4,-2,-1.
Hence, the real solutions to the equation are `x_1 = -2,x_2 = -1, x_3 = -4.`
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