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
.

By the remainder and factor theorems, if is a root then .

So you should find:

.

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

Once you have and , you have the polynomial.

Since is a root, is a factor.

Since is a root, is a factor.

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

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
- 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