# help solve 1

#### 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.

#### Prove It

MHF Helper
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.

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

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.

#### andy69

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

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
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.

#### andy69

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
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 $$\displaystyle \alpha$$.

#### andy69

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
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.

#### andy69

You know what $$\displaystyle a$$ and $$\displaystyle c$$ are now... Substitute them into your polynomial first.
4x^3+16x^2+44x-120 what do i do now?