The polynomials P (x)=x^3-x^2+4x and Q(x)= x^3+6x+10 leave the same remainder when divided by x-a .find the possible values of a and solve the inequality P (x) is greater than Q (x )

The Remainder Theorem states that the remainder is f(a). So, that means we have

a^3-a^2+4a = a^3+6a+10
a^2+2a+10 = 0

This has no real solutions.
I suspect a typo

As written, P is never greater than Q. See

http://www.wolframalpha.com/input/?i=plot+x%5E3-x%5E2%2B4x+,+x%5E3%2B6x%2B10

To find the possible values of \(a\) for which both polynomials \(P(x)\) and \(Q(x)\) leave the same remainder when divided by \(x-a\), we need to set up the equation using the remainder theorem.

The remainder theorem states that if a polynomial \(f(x)\) is divided by \(x-a\), then the remainder is equal to \(f(a)\).

So, for \(P(x)=x^3-x^2+4x\) to leave the same remainder as \(Q(x)=x^3+6x+10\) when divided by \(x-a\), we can set up the following equation:

\(P(a) = Q(a)\)

In this case, we have:

\(a^3-a^2+4a = a^3+6a+10\)

Simplifying this equation will enable us to find the possible values of \(a\).

\(a^3 - a^2 + 4a = a^3 + 6a + 10\)

This simplifies to:

\(-a^2 - 2a - 10 = 0\)

To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's solve it using the quadratic formula.

The quadratic formula states that for an equation of the form \(ax^2+bx+c=0\), the solutions for \(x\) are given by:

\[x = \frac{-b \pm \sqrt{{b^2 - 4ac}}}{2a}\]

In our equation, \(a=-1\), \(b=-2\), and \(c=-10\), so we can substitute these values in:

\[x = \frac{-(-2) \pm \sqrt{{(-2)^2 - 4(-1)(-10)}}}{2(-1)}\]

Simplifying further:

\[x = \frac{2 \pm \sqrt{{4 - 40}}}{-2}\]
\[x = \frac{2 \pm \sqrt{{-36}}}{-2}\]
\[x = \frac{2 \pm 6i}{-2}\]

The solutions are complex numbers. Therefore, there are no possible values of \(a\) for which both polynomials \(P(x)\) and \(Q(x)\) leave the same remainder when divided by \(x-a\).

Moving on to the second part of the question, to solve the inequality \(P(x) > Q(x)\), we need to compare the two polynomials.

\(P(x) = x^3 - x^2 + 4x\)

\(Q(x) = x^3 + 6x + 10\)

To find the values of \(x\) for which \(P(x) > Q(x)\), we can subtract \(Q(x)\) from \(P(x)\) and set the resulting expression greater than zero:

\(P(x) - Q(x) > 0\)

\((x^3 - x^2 + 4x) - (x^3 + 6x + 10) > 0\)

Simplifying this inequality:

\(-x^2 - 2x - 10 > 0\)

To solve this quadratic inequality, we can use different methods such as factoring or the graphical method. In this case, we will use the graphical method.

Plotting the graph of the quadratic function \(y = -x^2 - 2x - 10\), we need to determine the regions where the graph is above the x-axis (i.e., the values of \(x\) for which \(P(x) > Q(x)\)).

By analyzing the graph or factoring the quadratic equation to find the \(x\)-intercepts, we can determine the values of \(x\) that satisfy the inequality \(P(x) > Q(x)\).

To obtain the precise values, we can solve the quadratic equation using factoring, completing the square, or the quadratic formula.

Therefore, the values of \(x\) for which \(P(x) > Q(x)\) can be found by solving the quadratic equation \(P(x) - Q(x) > 0\) using the preferred method mentioned above.