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

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- May 19th 2010, 10:17 PMandy69help solve 1
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 - May 19th 2010, 10:47 PMProve It
$\displaystyle P(x) = ax^3 + 16x^2 + cx - 120$.

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. - May 19th 2010, 11:10 PMandy69
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. - May 20th 2010, 02:09 AMProve It
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. - May 20th 2010, 02:28 AMandy69
- May 20th 2010, 02:55 AMProve It
- May 20th 2010, 03:01 AMandy69
- May 20th 2010, 03:04 AMProve It
- May 20th 2010, 03:13 AMandy69