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Question 935110: I have been asked to show that (x+2) and (x-5) are factors of x^4+px^2+qx-40
Using synthetic division, I identified that for x+2;
Equation 1: 4p+16-2q-40
Equation 2: -25p+5q+585
However, if I were to subtract Equation 1 from 2; I do not have either P or Q fully nullified, there is a remainder left over.
For example, 4p-25p=-21p and 5q-2q=3q
However, in the case of 3p+27-q and -p-1+q
The -q and +q cancel each other out; leaving us only with the unknown P variable term. How would I remedy/rectify such a situation?
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! show that (x+2) and (x-5) are factors of x^4+px^2+qx-40
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Using synthetic division::
-2)....1....0....p....q....-40
.......1....8...p-16..q-2p+32..-40-2q+4p-64
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Equation:
4p - 2q + 24 = 0
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5)....1....0....p.........q........-40
......1....5....p+25.....q+5p+125..-40+5q + 25p + 625
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Equation::
25p + 5q +585 = 0
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System of equations in p and q::
2p - q = -12
5p + q = -117
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Add and solve for "p"::
7p =-129
p = -129/7
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Solve for "q":
2p - q = -12
-258/7 - q = -12
-258/7 - q = -84/7
q = -174/7
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Cheers,
Stan H.
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Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! 1) using synthetic division, we divide x^4+px^2+qx-40 by (x+2), note that the equation becomes x^4+0x^3+px^2+qx-40, then we have
-2 | 1 0 p q -40 |
we get the following cubic x^3-x^2+(4+p)x-8-2p+q where -8-2p+q is the constant
we also know that 16+4p-2q-40 = 0, therefore 4p-2q = 24, 2p-q=12
2) now use the cubic result and divide it by synthetic division
5 | 1 -2 4+p -8-2p+q
we get the following quadratic x^2+3x+19+p where 19+p is the constant
we also know that 95+5p-8-2p+q=0, 3p+q=-87
our two equations are
2p-q=12
3p+q=-87
solve first equation for p
p = (q/2)+6
substitute for p in second equation and solve for q, then p
3((q/2)+6)+q = -87
multiply both sides of = by 2
3q+36+2q=-174
5q = -210
q = -42
p = -15
our original equation is
x^4+px^2+qx-40
substitute for p and q
x^4-15x^2-42x-40
zeros for this equation using its graph
we see the graph crosses the x-axis at -2 and 5, these are the only zeros
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