$What am I doing wrong?$

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$What am I doing wrong?$

$x^2-5x+6$ and $x^3-6x^2+11x+a$ have a common factor. Find the possible value of a.

The way I tried to solve it was.

$f(x)=x^3-6x^2+11x+a$

$x^2-5x+6=(x-2)(x-3)$ so then $x=3$ or $x=2$

$f(3)=3^3-6\times 3^2+11\times3 +a=6+a=0 $
$-6 $ $-6 $
$=a=-6$

$f(2)=2^3-6\times 2^2+11\times2 +a=6+a=0$
$-6 $ $-6 $
$=a=-6$

So a = -6 but in my textbook, it says that it is incorrect and the actual answer is -66 or 18

Algebra

Mr Smurf

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OK, this looked to be an easy question but turned out to be a tricky one. I spent a couple of hours to find out what's going on.

Your approach is acorrect and makes sense. You could have done it much faster (less equations to solve) by multiplying the first equation by "x" and subtracting the second equation from it to get:

x^2−5x-a

Comparing with the first question, you'd have a=-6

Now, the issue is that the question asks for a common factor (i.e. one common factor) and not multiple common factors. See below:

x^2−5x+6=(x−2)(x−3)

x^3−6x^2+11x-6=(x-1)(x−2)(x−3)

These have two common factors: (x-1) and (x-3)

Hence, when you draw these, the two graphs have two intersections. See attached graphs.

If you solve the following for "a":

x^3−6x^2+11x-a=x^2−5x+6

First add -x^3 to both sides. Then, add 6x^2 to both sides. and finally add -11x to both sides, you'll get:

a=-x^3+7x^2-16x-6

This have infinite solutions for "a", 18 (when x=-1) and -66 (when x=6) being two of those.

x^3−6x^2+11x+18=(x+1)(x^2-7x+18)

x^3−6x^2+11x-66=(x-6)(x^2+11)

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Kav10

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