# help solve 1

• May 19th 2010, 10:17 PM
andy69
help 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.

• May 19th 2010, 10:47 PM
Prove It
Quote:

Originally Posted by andy69
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.

$P(x) = ax^3 + 16x^2 + cx - 120$.

By the remainder and factor theorems, if $a$ is a root then $P(a) = 0$.

So you should find:

$P(-2) = a(-2)^3 + 16(-2)^2 - 2c - 120 = 0$

$P(3) = a(3)^3 + 16(3)^2 + 3c - 120 = 0$.

After you have simplified these, you have 2 equations in 2 unknowns. Solve them simultaneously for $a$ and $b$.

Once you have $a$ and $b$, you have the polynomial.

Since $-2$ is a root, $x + 2$ is a factor.

Since $3$ is a root, $x - 3$ is a factor.

Long divide your polynomials by these factors to find the third factor. From there you can find the final root $\alpha$.

If you have any more trouble, please show your working and exactly where you are stuck.
• May 19th 2010, 11:10 PM
andy69
Quote:

Originally Posted by Prove It
$P(x) = ax^3 + 16x^2 + cx - 120$.

By the remainder and factor theorems, if $a$ is a root then $P(a) = 0$.

So you should find:

$P(-2) = a(-2)^3 + 16(-2)^2 - 2c - 120 = 0$

$P(3) = a(3)^3 + 16(3)^2 + 3c - 120 = 0$.

After you have simplified these, you have 2 equations in 2 unknowns. Solve them simultaneously for $a$ and $b$.

Once you have $a$ and $b$, you have the polynomial.

Since $-2$ is a root, $x + 2$ is a factor.

Since $3$ is a root, $x - 3$ is a factor.

Long divide your polynomials by these factors to find the third factor. From there you can find the final root $\alpha$.

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.
• May 20th 2010, 02:09 AM
Prove It
Quote:

Originally Posted by andy69
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

$24a + 6c + 168 = 0$

but you are incorrect with equation 2.

You should have gotten:

$27a + 144 + 3c - 120 = 0$

$27a + 3c + 24= 0$

$54a + 6c + 48 = 0$.

So the two equations are:

$24a + 6c + 168 = 0$

$54a + 6c + 48 = 0$.

Go from here.
• May 20th 2010, 02:28 AM
andy69
Quote:

Originally Posted by Prove It
Incorrect.

You are right that you should transform equation 1 into

$24a + 6c + 168 = 0$

but you are incorrect with equation 2.

You should have gotten:

$27a + 144 + 3c - 120 = 0$

$27a + 3c + 24= 0$

$54a + 6c + 48 = 0$.

So the two equations are:

$24a + 6c + 168 = 0$

$54a + 6c + 48 = 0$.

Go from here.

24a + 6c + 168 = 0(1)
54a + 6c + 48 = 0(2)
(2)-(1)
30a-120=0
30a=120
a=120/30
a=4
sub a=4 into (1)
96+6c+168=0
6c=-168-96
6c=-264
c=-264/6
c=-44

is this working out right
• May 20th 2010, 02:55 AM
Prove It
Quote:

Originally Posted by andy69
24a + 6c + 168 = 0(1)
54a + 6c + 48 = 0(2)
(2)-(1)
30a-120=0
30a=120
a=120/30
a=4
sub a=4 into (1)
96+6c+168=0
6c=-168-96
6c=-264
c=-264/6
c=-44

is this working out right

Yes, well done. (Clapping)

Now long divide by $x + 2$ and $x - 3$. From the third factor you'll be able to find $\alpha$.
• May 20th 2010, 03:01 AM
andy69
Quote:

Originally Posted by Prove It
Yes, well done. (Clapping)

Now long divide by $x + 2$ and $x - 3$. From the third factor you'll be able to find $\alpha$.

so do i divide these 2 factors with p(x)=ax3 +16x2+cx – 120
• May 20th 2010, 03:04 AM
Prove It
Quote:

Originally Posted by andy69
so do i divide these 2 factors with p(x)=ax3 +16x2+cx – 120

You know what $a$ and $c$ are now... Substitute them into your polynomial first.
• May 20th 2010, 03:13 AM
andy69
Quote:

Originally Posted by Prove It
You know what $a$ and $c$ are now... Substitute them into your polynomial first.

4x^3+16x^2+44x-120 what do i do now?