The polynomial p(x)is given by p(x)=ax3 +16x2+cx – 120, where a and c are constants. The three zeros of p(x)are –2, 3 and α. Find the value of α.
with working out please
$\displaystyle P(x) = ax^3 + 16x^2 + cx - 120$.
By the remainder and factor theorems, if $\displaystyle a$ is a root then $\displaystyle P(a) = 0$.
So you should find:
$\displaystyle P(-2) = a(-2)^3 + 16(-2)^2 - 2c - 120 = 0$
$\displaystyle P(3) = a(3)^3 + 16(3)^2 + 3c - 120 = 0$.
After you have simplified these, you have 2 equations in 2 unknowns. Solve them simultaneously for $\displaystyle a$ and $\displaystyle b$.
Once you have $\displaystyle a$ and $\displaystyle b$, you have the polynomial.
Since $\displaystyle -2$ is a root, $\displaystyle x + 2$ is a factor.
Since $\displaystyle 3$ is a root, $\displaystyle x - 3$ is a factor.
Long divide your polynomials by these factors to find the third factor. From there you can find the final root $\displaystyle \alpha$.
If you have any more trouble, please show your working and exactly where you are stuck.
ok is this right um i subbed -2 and 3 into the equations
and got -8a+64-2c-120=0(1) and 27a+144+3c-120=0(2)
then i did -8a+64-2c-120(times by -3) and 27a+144+3c-120(times by 2)
which is 24a+6c+168=0(3) and 54a+6c+44=0(4)
then i did (4)-(3)
which is 30a-124=0
30a=124
a=124 divide by 30
=4.13
sub a=4.13 into (4)
can you tell me if this is right so far.
Incorrect.
You are right that you should transform equation 1 into
$\displaystyle 24a + 6c + 168 = 0$
but you are incorrect with equation 2.
You should have gotten:
$\displaystyle 27a + 144 + 3c - 120 = 0$
$\displaystyle 27a + 3c + 24= 0$
$\displaystyle 54a + 6c + 48 = 0$.
So the two equations are:
$\displaystyle 24a + 6c + 168 = 0$
$\displaystyle 54a + 6c + 48 = 0$.
Go from here.