Use synthetic division and the given factor to completely factor the polynomial function. x^3+7x^2+7x-15=(x+5)

Question

Could you help me understand how to solve this?

Answer 1

Hi Ruby S.,

"Synthetic division" is just a fancy name for the process of dividing one polynomial (dividend) by another polynomial (divisor), to obtain a polynomial (quotient) as the answer.

If you are careful about how you write your polynomials, you don't even need to write your x^3, x^2, x, etc. as part of this process, although you still could if you wanted. They are understood to be present (==implicit), even if you don't write them. Instead, you just write the leading coefficients of each term of the polynomial, making sure to include a "0" if there is a missing term in a polynomial.

You perform "long division" on the coefficient sets as follows:

First, write the coefficients of the dividend you are starting with. So, here you have:

1 7 7 -15 . Note that I've included spaces between the coefficients, to help with the bookkeeping!

Then, write your long division symbol to the left of and above that number set: looks sort of like:

___________________

| 1 7 7 -15

Next, write your coefficient set for your divisor, to the left of all that, just as you would have done for long-dividing two numbers:

. ______________

1 5 | 1 7 7 -15

Next, start your division process. For each step, you are "processing" as many "places" as there are places in the divisor. This is just like doing long division on two numbers, except that each of the numbers present represents a coefficient in a polynomial, meaning that they can be any integer, including negative, and even greater than 10 or less than -10! (If they are not integers, it's probably easier to multiply that entire polynomial up to integer coefficients.)

Then, you start mentally checking, how many times can that divisor "number set" go into the corresponding length portion of the start of the dividend number set. Here, you notice that one times 1 5 is < 1 7 but that twice that, or 3 0 would be > 1 7 . So, the first "digit" of your quotient is a 1. Write that above the first 7 in the dividend, just as you would do for regular long division:

_____1__________

1 5 | 1 7 7 -15

-- (1 5)

____________

2 7 -15

Note that I have written the 1 5 as a subtracted "number set", performed the subtraction, written the result, and also carried the remaining portions of the dividend number set down. Unlike in regular long division, the subtraction is done by places without borrowing. If the result is a negative number for any place, so be it! There are some things to appreciate here: any coefficient in this expression may be a negative number, and you need to be a careful bookkeeper to deal with that! For example, if that divisor had negative values anywhere, when you subtract by it or a multiple of it, that place would have a double negative sign on it, so it would be a + operation. So if the divisor above were (1 -5), that first subtraction would have been -- (1 -5) and the result would have been 12 7 -15 at that stage of the division.

So, you carry on throughout the entire length of the dividend, in the same way.

Your final result should look like:

____1 ___2 ___-3___

1 5 | 1 7 7 -15

-- (1 5)

________________

2 7 -15

-- (2 10)

_________________

-3 -15

-- ( -3 -15)

___________________

0

The final "0" show you that the division was exact, i.e. there was no remainder!

And the (1 2 -3 ) of the quotient tells you that the other factor was (1x^2 + 2x -3).

OK?

-- Cheers, -- Mr. d.

Answer 2

To divide by (x + 5), we use synthetic division with the associated zero of "x + 5" which is x = -5

So to put back together the resulting polynomial quotient, we use the resulting numbers as the coefficients:

or 1x² + 2x - 3

So x³ + 7x² + 7x - 15 = (x + 5)(x² + 2x - 3) and, using standard quadratic factoring techniques, we can factor x² + 2x - 3 into (x + 3)(x - 1) meaning x³ + 7x² + 7x - 15 = (x + 5)(x + 3)(x - 1)