# 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

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

Mr Smurf

32

## Answer

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Kav10

1.9K

The answer is accepted.

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