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

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

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.

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.

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Incorrect.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.

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)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.

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)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

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 – 120Yes, well done. (Clapping)

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

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

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