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# System prepended metadata

title: PID-714

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We can use the factor theorem and the remainder theorem to solve this problem.

**(1)** To find the values of $a$ and $b$, we use the factor theorem. Since $x-2$ is a factor of $p(x)=ax^3-9x^2+bx-6$, we know that $p(2) = 0$. Substituting $x=2$ into the polynomial $p(x)$, we get:
$$p(2) = a(2)^3 - 9(2)^2 + b(2) - 6 = 0$$

Simplifying this equation, we get:
$$8a + 2b = 42.$$

Dividing throughout by 2, we have
$$4a+b=21.\tag1$$

Next, we use the remainder theorem. Since the polynomial has a remainder of 66 when divided by $x-3$, we know that $p(3) = 66$. Substituting $x=3$ into the polynomial, we get:
$$p(3) = a(3)^3 - 9(3)^2 + b(3) - 6 = 66$$

Simplifying this equation, we get:
$$27a + 3b = 153$$

Dividing throughout by 3, we have
$$9a+b=51.\tag2$$

Subtracting equation (1) from equation (2), we have
$$5a=30$$

$$\implies a=6.~~~~~~~~~$$

Substituting the value $a=6$ in equation (1), we have
$$4(6) +b=21.$$

$$b=21-24=-3.$$
Thus, $a=6$ and $b=-3$

**(2)** Using the values $a=6$ and $b=-3$, we have:
$$p(x) = 6x^3-9x^2-3x-6.$$

Dividing $p(x)= 6x^3-9x^2-3x-6$ by $x-2$ by long division method as follows:

![](https://hackmd.io/_uploads/rJU0tPRTh.jpg)

Thus, we can express the polynomial $p(x)$ as follows:
$$p(x)=(x-2)(6x^2+3x+3).$$

**(3)** To show that the equation $p(x) = 0$ has only one real solution, we show that the quadratic factor $q(x)=6x^2+3x+3$ (attained in **(2)**) has no real root. For this, we show the discriminant of the quadratic equation $q(x)=6x^2+3x+3$ is negative. The discriminant is:
$$\Delta = b^2 - 4ac = 3^2-4(6)(3)=-63$$

Since the discriminant is negative, the quadratic factor $q(x)=6x^2+3x+3$ has two distinct imaginary roots. Therefore, the only real solution of $p(x) = 0$ is $x=2$, which is the root corresponding to the factor $x-2$.