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# If ${x^3} + m{x^2} + nx + 6$ has $(x - 2)$ as factor and leaves a remainder $3$ when divided by $(x - 3)$ find the values of $m,n$ .$A)m = 2,n = 2$$B)m = 2,n = - 2$$C)m = - 2,n = 1$$D)m = - 3,n = - 1$

Last updated date: 18th Sep 2024
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Hint: First we have to define what the terms we need to solve the problem are.
Since in the given question we first need to find the factors and then only we are able to find $m,n$because without knowing the factors of the given problem it is nowhere to find.
The first factor is $(x - 2)$

As we know that to find the m, n in the polynomial equation of ${x^3} + m{x^2} + nx + 6$ and this polynomial fact has the factor of $(x - 2)$ and also it will leave the remainder $3$ when we divisible by $(x - 3)$
Now $(x - 2)$ is the factor that means we approach it to zero that find the value of $x$ $(x - 2) = 0 \Rightarrow x = 2$ so then the value of $x$ is two as know, we are just going to substitute the value of $x$so that we can find the polynomial simplification to $m,n$. That is
Take $f(x) = {x^3} + m{x^2} + nx + 6$ now put $x = 2$ as per above solution we get $f(2) = {2^3} + m.{2^2} + n.2 + 6 = 14 + 4m + 2n$(approaching and simplifying)
Now let us take to function of polynomial to zero so that we can able to simply the above equation
$\Rightarrow 7 + 2m + n = 0$(Taking two as common and eliminated as per right hand side is zero) ---$\left( 1 \right)$
Similar now $(x - 3) = 0$(solve as per same above) we get $x = 3$
Substitute in polynomial factor and we get $f(3) = {3^3} + m.{3^2} + n.3 + 6-3$(as per remainder zero and solve further we get, below) $10 + 3m + n = 0$ ---$\left( 2 \right)$
Hence, we have two equations using the given factors and polynomial, so that we are going to solve using elimination method, solve equation one and two we get;
First find the one value $m = - 3$ and apply it in the equations we get then $n = - 1$

So, the correct answer is “Option D”.

Note: We can use elimination or substitution method for the two know polynomial functions each will occur the same answers, so no problem of using the different formula; also, for all other options there is no such way possible; $A)m = 2,n = 2$,$B)m = 2,n = - 2$,$C)m = - 2,n = 1$,
We are also able to check by substituting the m, n in the polynomial so all other options are eliminated because it will not be going to be satisfied hence D is correct.