Question

h(x) = x³ + ax² + bx + c the function h is defined above where a, b and c are integer constants if the zero of the function are -5 , 6 , and 7 what is the value of c?

Answer 1

The zeros of the given function are -5, 6 and 7.

$\begin{gathered} h(x)=x^3+ax^2+bx+c \\ \text{When x=-5,} \\ h(x)=(-5)^3+a(-5)^2+b(-5)+c=0 \end{gathered}$$\begin{gathered} -125+25a-5b+c=0 \\ 25a-5b+c=125\ldots.eqn(1)_{} \end{gathered}$$\begin{gathered} \text{When x=6,} \\ (6)^3+a(6^2)+6b+c=0 \\ 216+36a+6b+c=0 \\ 36a+6b+c=-216\ldots..eqn(2) \end{gathered}$$\begin{gathered} \text{When x=7,} \\ h(x)=7^3+a(7^2)+7b+c=0 \\ 343+49a+7b+c=0 \\ 49a+7b+c=-343\ldots..eqn(3) \end{gathered}$

To get the value of c:

We solve simultaneously the 3 equations;

From eqn(1),

$\begin{gathered} b=\frac{25a+c-125}{5}\ldots\text{eqn}(4) \\ \text{From eqn(2),} \\ b=\frac{-36a-c-216}{6}\ldots.eqn(5) \\ \text{From eqn(3),} \\ b=\frac{-49a-c-343}{7}\ldots.eqn(6) \end{gathered}$

So, equating equations (4) and (5), we get,

$\begin{gathered} \frac{25a-125+c}{5}=\frac{-36a-c-216}{6} \\ \text{Cross multiplying we get,} \\ 150a-750+6c=-180a-5c-1080 \\ 330a+11c=-330\ldots.eqn(7) \end{gathered}$

Similarly, equating equations (4) and (6), we get,

$\begin{gathered} \frac{25a+c-125}{5}=\frac{-49a-c-343}{7} \\ \text{Cross multiplying we get,} \\ 175a+7c-875=-245a-5c-1715 \\ 420a+12c=-840\ldots.eqn(8) \end{gathered}$

Solving equations (7) and (8) simultaneously we get,

$\begin{gathered} 330a+11c=-330\ldots.eqn(7) \\ 420a+12c=-840\ldots.eqn(8) \\ 14(\text{eqn}7);\text{ 14(330a+11c=-330) ; 4620a+154c=-4620 }\ldots eqn(9) \\ 11(\text{eqn}8);\text{ 11(420a+12c=-840) ; 4620a+132c=-9240 }\ldots eqn(10) \\ \text{Subtracting eqn(10) from eqn(9), we get,} \\ 22c=4620 \\ \text{Dividing both sides by 22, we get,} \\ c=\frac{4620}{22}=210 \end{gathered}$

Solution: The value of c is 210.