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Answer 1 · 2017-09-05T10:23:16

There are two possible values:

$m = - \frac{3}{2}$ $m=-3/2$ or $m = 2$ $m= 2$

We use the remainder theorem:

The remainder of the division of a polynomial $f (x)$ $f(x)$ by a linear factor $x - a$ $x-a$ is $f (a)$ $f(a)$

Consider the first polynomial:

$P (x) = x^{3} + 4 x^{2} - 2 x + 1$ $P(x)=x^3 +4x^2 - 2x +1$

Then by the Remainder Theorem, if we divide $P (x)$ $P(x)$ by $x - m$ $x-m$ then the remainder, $r_{p}$ $r_p$ , is:

$r_{p} = P (m) = m^{3} + 4 m^{2} - 2 m + 1$ $r_p = P(m) = m^3 +4m^2 - 2m +1$

Similarly for the second polynomial:

$Q (x) = x^{3} + 2 x^{2} - x + 7$ $Q(x) = x^3 +2x^2 - x + 7$

If we divide $Q (x)$ $Q(x)$ by $x - m$ $x-m$ then the remainder, $r_{q}$ $r_q$ , is:

$r_{q} = Q (m) = m^{3} + 2 m^{2} - m + 7$ $r_q = Q(m) = m^3 +2m^2 - m + 7$

We are given that the remainders are the same:

$:. r_p = r_q$
$:. m^3 +4m^2 - 2m +1 = m^3 +2m^2 - m + 7$
$:. 2m^2 - m -6 = 0$
$:. (m-2)(2m+3) = 0$
$:. m=-3/2, 2$