You can put this solution on YOUR website! Simplify, and then put into general form; try to factor by grouping if this is possible and a way can be efficiently found; otherwise, use Rational Roots Theorem.
Product of the two trinomials using lattice multiplication arrangement:
x^2 7x 12
______________________________________
|
x^2 | 2x^4 7x^3 12x^2
|
-3x | -3x^3 -21x^2 -36x
|
2 | 2x^2 14x 24
Use synthetic division to check possible roots of -1,-2,-4,-7,1,2,4,7 for this degree-four polynomial.
You will find that no rational zeros will be found; you will see using a graphing tool that no real zeros occur. Original expression is not factorable in real numbers.
Yes it is factorable. The other tutor thought
because it has no rational zeros (roots) that it
is not factorable. That is not the case. A 4th
degree polynomial could factor into the product
of two quadratic trinomials both which have
irrational zeros (roots).
Here is the correct approach:
(x-1)(x-2)(x+3)(x+4)+4
If we were going to multiply that out, we would
start by multiplying two pairs of those binomials
by FOIL. We have a choice of
1. multiplying the 1st by the 2nd and the 3rd by
the 4th, or
2. multiplying the 1st by the 3rd and the 2nd by
the 4th, or
3. multiplying the 1st by the 4th and the 2nd by
the 3rd, or
We investigate the 3 possibilities.
If we multiply the 1st by the 2nd and the 3rd by
the 4th the middle terms would be -3x and 7x.
If we multiply the 1st by the 3rd and the 2nd by
the 4th, the middle terms would be 2x and 2x.
Both the same.
If we multiply the 1st by the 4th and the 2nd by
the 3rd the middle terms would be 3x and x.
The 2nd way is the best plan for we get the
same middle term.
Rearrange the factors so that the ones we multiply
will be side by side:
(x-1)(x+3)(x-2)(x+4)+4
(x²+2x-3)(x²+2x-8)+4
Let u = x²+2x
(u-3)(u-8)+4
u²-11u+24+4
u²-11u+28
(u-7)(u-4)
Replace u by x²+2x
(x²+2x-7)(x²+2x-4)
Edwin
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