Download A polynomial of degree n (in one variable, with real coefficients) is

A polynomial of degree n (in one variable, with real coefficients) is an
expression of the form:
an xn + an−1 xn−1 + an−2 xn−2 + · · · + a2 x2 + a1 x + a0
where an , an−1 , an−2 , · · · a2 , a1 , a0 are real numbers.
Example:
3x4 − 2x2 + 1 is a polynomial of degree 4.
−x10 + 7x5 − 2x3 + x − 5 is a polynomial of degree 10.
−2x is a polynomial of degree 1.
The constant expression 2 is a polynomial of degree 0.
Notice that a linear expression is a polynomial of degree 1.
The degree of a polynomial is the highest power of x whose coefficient is not 0.
By convention, a polynomial is always written in decreasing powers of x.
The coefficient of the highest power of x in a polynomial is the leading coefficient.
In the above example, the leading coefficient of p is 3.
The leading coefficient of f is −1.
The coefficient of a polynomial is understood to be 0 if the term is not shown.
A polynomial of one term is a monomial.
A polynomial of two terms is a binomial.
A polynomial of three terms is a trinomial.
A polynomial of degree 1 is a linear expression.
A polynomial of degree 2 is a quadratic expression.
A polynomial of degree 3 is a cubic expression.
A polynomial of degree 4 is a quartic expression.
A polynomial of degree 5 is a quintic expression.
To add two polynomials, add their terms by collect like terms.
E.g.
(x3 + 2x2 − 3x + 4) + (−4x4 + 2x3 − 4x2 + x + 1) = −4x4 + 3x3 − 2x2 − 2x + 5
To subtract two polynomials, take the negatives of the second expression and
add.
Example:
(2x4 − 2x3 + 3x − 3) − (5x4 − 3x3 + x2 − 7x + 6)
= 2x4 − 2x3 + 3x − 3 − 5x4 + 3x3 − x2 + 7x − 6
= −3x4 + x3 − x2 + 10x − 9
To multiply two polynomials, treat the first polynomial as if it is one single
term, and apply the distributive property to distribute the first polynomial to
each of the terms of the second polynomial, then multiply each term of the
second polynomial to each of the terms in the first polynomial by the distributive
property again.
E.g.
(3x2 + 2x − 1)(2x − 3)
treat (3x2 + 2x − 1) as a single term and distribute this to (2x − 3) by applying
the distributive property.
= (3x2 + 2x − 1)(2x) − (3x2 + 2x − 1)(3)
multiply 2x to each of the terms of the first polynomial, and multiply −3 to
each of the terms in the first polynomial using the distributive property again.
Remember to distribute the negative sign as well.
6x3 + 4x2 − 2x − 9x2 − 6x + 3
collect like terms to simplify.
6x3 − 5x2 − 8x + 3
There is another method of multiplying polynomials, which is to multiply each
term of the first polynomial with each term of the second polynomial:
(3x2 + 2x − 1)(2x − 3) = 6x3 − 9x2 + 4x2 − 6x − 2x + 3
(each term of the first polynomial is multiplied to each terms of the second
polynomial. Notice that the negative sign is also multiplied.
= 6x3 − 5x2 − 8x + 3
In algebra we often need to multiply two binomials (usually of degree 1), using
the method of multiplying each term of the first polynomial to each term of the
second polynomial we have:
(x + 2)(−x − 4) = −x2 − 4x − 2x − 8
The terms that are being multiplied together can be viewed as:
The First terms ((x)(−x)) because they are the first terms of the two polynomials.
The Outer terms ((x)(−4)) because they are the terms on the outside when the
expression is written horizontally.
The Inner terms ((2)(−x)) because they are the terms on the inside when the
expression is written horizontally.
The Last terms ((2)(−4)) because they are the last terms of the two polynomials.
This is called the FOIL method.
Some special products:
(a + b)(a − b) = a2 − ab + ab − b2 = a2 − b2 (difference of two squares)
(a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 (square of a sum)
(a − b)2 = (a − b)(a − b) = a2 − ab − ab + b2 = a2 − 2ab + b2 (square of a sum)
Factoring
Factoring the greatest common factor:
E.g. (x2 − 4x) = x(x − 4) This is the reverse process of the distributive property.
x is the GCF of the two terms, and it can be factored out.
(−12x4 + 6x2 + 4x) = 2x(−6x3 + 3x + 2)
(4x+2) = 2(2x+1) Notice in this example that, if one of the term in an expression
is the GCF, when that term is factored out, what remains behind is the number
1.
Factoring out the negative sign
E.g. (3x − 4) = −1(−3x + 4) = −(4 − 3x)
(−2x3 + 4x2 − 2x) = −2x(x2 − 2x + 1)
Factor by Grouping:
Factor by Grouping may be applied if an expression to be factored has an even
number of terms (4, 6, 8...) and the terms do not share a common factor, but
some of the terms have common factors. In this case, group the terms into 2
groups of equal number of terms (2 each if there are 4 terms), where at least one
of the groups should have a common factor.
E.g. (x3 + 3x2 + 4x + 12) = (x3 + 3x2 ) + (4x + 12) = x2 (x + 3) + 4(x + 3) =
(x2 + 4)(x + 3)
E.g. ax − 4ay + 3bx − 12by = (ax + 3bx) − (4ay + 12by) notice that we used the
method of factoring out the negative sign.
= x(a + 3b) − 4y(a + 3b) = (x − 4y)(a + 3b)
Factoring trinomial of the form: ax2 + bx + c
If a = 1
We use trial and error method:
E.g.
x2 + 5x − 6
We are looking for two numbers that multiply to −6 and adds up to 5, Since
(6)(−1) = −6 and 6 + −1 = 5, 6 and −1 are our choices, we have:
x2 + 5x − 6 = (x + 6)(x − 1)
x2 + 7x + 12
Two numbers that multiply to 12 and adds up to 7, so 3 and 4 works, we have:
x2 + 7x + 12 = (x + 3)(x + 4)
Factoring quadratic when a 6= 1:
6x2 + 13x − 8
The trial and error method still works, but need to be applied more carefully.
You choose a factoring of 6, say 6 and 1, to test if it works:
6x2 + 13x − 8 = (6x + a)(x + b)
You are looking for a and b such that ab = −8, and 6b+a = 13 (6b+a contributes
to the middle term). There is no two numbers a, b that would work. But we are
not done, we may try 3 and 2 since (3)(2) = 6.
6x2 + 13x − 8 = (3x + a)(2x + b)
We are now looking for a and b such that ab = −8 and 3b + 2a = 13. Notice that
a = 8 and b = −1 works:
6x2 + 13x − 8 = (3x + 8)(2x − 1)
AC-method
A more systematic method is to multiply a with c, then look for two numbers
that would multiply to ac and add up to b, then split the middle term bx into
the sum of that two numbers, then factor by grouping:
E.g.
12x2 + x − 6
In this example, a = 12 and c = −6. So ac = (12)(−6) = −72. We are looking
for two numbers that would multiply to −72 and adds up to the middle term
(b = 1). In this example, 9 and −8 work, so we split the middle term like this:
12x2 + x − 6 = 12x2 + 9x + −8x − 6
Then factor by grouping:
12x2 + 9x + −8x − 6 = (12x2 + 9x) − (8x + 6) = 3x(4x + 3) − 2(4x + 3) =
(3x − 2)(4x + 3)
The sum or difference of two cubes can be factored:
a3 + b3 = (a + b)(a2 − ab + b2 )
a3 − b3 = (a − b)(a2 + ab + b2 )
Solving Equations by Factoring:
The Zero Factor Property of Real Numbers:
If ab = 0, then a = 0 or b = 0 (or both = 0).
To solve the equation:
x2 + 4x + 3 = 0, we cannot try to isolate the variable by doing just algebraic
operations. Instead, we factor the left hand side:
(x + 3)(x + 1) = 0
Since (x + 3) and (x + 1) multiplies to 0, the zero factor property tells us that
one of them must be 0, we have:
x + 3 = 0 or x + 1 = 0
Solving the two equations we get:
x = −3 or x = −1
In trying to solve an equation by factoring, one of the sides of the equation
(usually the right hand side) must be equal to 0.
E.g.
x2 − 6x + 8 = 15
If you try to solve this equation by factoring, you must first move the 15 to the
other side to make the right hand side 0. (If you try to factor the expression on
the left immediately, you would be wrong).
x2 − 6x − 7 = 0
(x − 7)(x + 1) = 0
x − 7 = 0 or x + 1 = 0
x = 7 or x = −1
E.g
4x2 − 12x − 7 = 0
(2x − 7)(2x + 1) = 0
2x − 7 = 0 or 2x + 1 = 0
7
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x = or x = −
2
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