help solve 1

May 2010
39
0
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
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
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.
 
May 2010
39
0
\(\displaystyle P(x) = ax^3 + 16x^2 + cx - 120\).


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.
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,897
5,001
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.
 
May 2010
39
0
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.
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
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
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 \(\displaystyle x + 2\) and \(\displaystyle x - 3\). From the third factor you'll be able to find \(\displaystyle \alpha\).
 
May 2010
39
0
Yes, well done. (Clapping)

Now long divide by \(\displaystyle x + 2\) and \(\displaystyle x - 3\). From the third factor you'll be able to find \(\displaystyle \alpha\).
so do i divide these 2 factors with p(x)=ax3 +16x2+cx – 120
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
so do i divide these 2 factors with p(x)=ax3 +16x2+cx – 120
You know what \(\displaystyle a\) and \(\displaystyle c\) are now... Substitute them into your polynomial first.