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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 fol lowing 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 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 de termine 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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