# Preparation for Math Chapter 4

**1- Synthetic Division **

If a polynomial p (x) is divided by (x – c) we can apply synthetic division:

Let’s divide x^{3} – 4x^{2} – 5 by x – 3

Step 1: Write the dividend in descending powers of x. Copy the coefficients;
write a zero

for any missing power .

1 -4 0 -5

Step 2: Create the following structure:

First row, we place: 1 -4 0 -5

Second row, to the left of the symbol , we place 3 since the divisor is x – 3.

Step 3: Bring the 1 down two rows , and enter it in row 3.

Step 4: Multiply the 1, in row 3, by 3, and place the result 3, in row 2 underneath the -4.

Step 5: Add the entries in row 1 and row 2 (-4 + 3 = -1) and enter the sum in row 3.

Step 6: Multiply the -1, in row 3, by 3, and place the result -3, in row 2 underneath the 0.

Step 7: Add the entries in row 1 and row 2 (0 + -3 = -3) and enter the sum in row 3.

Step 8: Multiply the -3, in row 3, by 3, and place the result -9, in row 2 underneath the -5.

Step 9: add the entries in row 1 and row 2 (-5 + -9 = -14) and enter the sum in row 3.

End of the process.

The numbers 1 -1 and -3 are the coefficients of a second degree polynomial that
is the

quotient (x^{2} – x – 3), the number -14 is the remainder, so

x^{3} - 4x^{2} – 5 = (x - 3)(x^{2} – x – 3) + (-14)

**Example:**

Use synthetic division to divide (2x^{5} + 5x^{4} – 2x^{3} + 2x^{2} – 2x + 3) by (x + 3)

The divisor is x + 3 = x – (-3), so we place -3 to the left of the symbol.

Remember, we add each entry in row 1 to the corresponding entry in row 2 and the
sum

is place in row 3.

The remainder is 0, the quotient is the polynomial 2x^{4} –
x^{3} + x^{2} – x + 1, so

2x^{5} + 5x^{4} – 2x^{3} + 2x^{2} – 2x + 3) = (x + 3)( 2x^{4} – x^{3} + x^{2} – x + 1)

**2- Roots or Zeros of a Polynomial
**Given a polynomial P(x), we say that a number c is a root or zero of the
polynomial if

P(c) = 0.

The Factor Theorem :

The Factor Theorem :

The number c is a root or zero of the polynomial P(x) if and only if (x – c) is a factor of

P(x). (P(x) = (x – c) Q(x) where degree Q(x) equals degree P(x) minus 1)

**Example:**

Show that x – 1 is a factor of P(x) = 2x

^{3}– x

^{2}+ 2x – 3.

Applying the theorem, we have to show that P(1) = 0.

P(1) = 2(1)

^{3}– (1)

^{2}+ 2(1) – 3 = 2 – 1 + 2 – 3 = 0

Then P(x) = 2x

^{3}– x

^{2}+ 2x – 3 = (x -1)Q(x).

We can find Q(x) using synthetic division,

Q(x) = 2x^{2} – x + 3 and P(x) = 2x^{3} – x^{2} + 2x – 3 = (x -1)(
2x^{2} – x + 3).

**3- Quadratic Equation **

A quadratic equation has the form : ax^{2} + bx + c = 0 where a, b, and c are real
numbers.

We can use quadratic formula to find the roots or zeros,

b^{2} – 4ac is called the **discriminat.**

If b^{2} – 4ac > 0, there are two unequal real number solutions.

If b^{2} – 4ac = 0 there is a repeated solution, root of multiplicity 2.

If b^{2} – 4ac < 0 there are no real solutions, two distinct complex number
solutions.

**4- The Rational Zeros Theorem**

Let be a polynomial with integer

coefficients.

A rational number p/q in lowest terms , is a rational zero of P(x) if p is a
factor of and q

is a factor of .

**Example:**

List the potential rational zeros of P(x) = 2x^{3} + 11x^{2} – 7x – 6.

Notice that the coefficients of P(x) are integer numbers.

Since = 2 and
= -6, then p is a factor of -6 and q is a factor of 2:

p : ±1, ±2, ±3, ±6 (factors of -6)

q: ±1, ±2 (factors of 2)

Now, we form all possible ratios p/q: ±1, ±2, ±3, ±6, ±1/2, ±3/2. There are 12
possible

candidates. So if P(x) has a rational zero, it is one of those listed. It may be
the case that

P(x) does not have any rational zero.

By substitution , we can check each potential zero, to determine whether it is
indeed a

zero.

Let’s start with 1, P(1) = 2(1)^{3} + 11(1)^{2} – 7(1) – 6 = 0. It is a zero, then (x
– 1) is a factor

of P(x).

We can use synthetic division to factor P(x)

Then, P(x) = (x -1)(2x^{2} + 13x + 6) = (x – 1)(x + 6)(x +
½), using quadratic formula, so 1,

-6, and -1/2 are the zeros of P(x).

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