2x 2 Xy 3y 2
3.1 Prime number Factors for Polynomials
The student should begin this chapter with a review of the thought of factoring integers. which was discussed in Chapter 1.
A polynomial P is said to he a factor or divisor of a polynomial R if at that place exists a polynomial Q such that
R=PQ
Notation thatQ is besides a divisor ofR.
In this chapter we will agree that our polynomials are to have only integral coefficients. For instance,
10^two-1=(ten-1)(x+1)
and thereforex^two-ane has factors x-1 and ten+1. But, fifty-fifty though
10^2-6=(x-root(6))(10+root(6))
we do non consider 10-root(6) and .ten+√6 as factors ofx^2-6 since they fail tn have integral coefficients. A given polynomial with integral coefficients is said to he prime if it has no factors other than plus or minus one and plus or minus itself, bailiwick to the in a higher place restrictions. A polynomial is said to exist factored completely when information technology is expressed as a product of prime number factors.
When nosotros are obtaining factors of polynomials we must make allowances for changes in sign. Since
ten^ii-1=(10-1)(ten+ane)
=(-one)(-ten+i)(x+1)
and we do not desire to distinguish between these factorization, we volition agree that if the factors differ only by a multiple of -one, they volition exist considered to be equivalent. With this understanding the following is true: Every polynomial tin be expressed uniquely as the product of prime factors apart grade the order in which they are written and discipline to little changes in sign. This is known as the Unique Factorization Theorem for Polynomials.
The problem of finding all of the prime factors of a given polynomial is, in general, a hard one. In this chapter nosotros consider techniques for finding factors of sure types of polynomials. In guild to find the prime factors of a given polynomial we must be able to tell whether or non a polynomial is prime number. Ane type of polynomial that is always prime is a polynomial of first degree, ax + b, for which the only mutual factors of a andb are +-1. For example,3x+4 is prime number since 3 and 4 take only +-i equally common factors, while 4x + 6 is not prime since two is a common factor of 4 and six. In fact, 4x +half dozen = 2(2x + 3).
Using long segmentation nosotros can determine if a given polynomial is a factor of some other polynomial. This is the case if and only if the residual is naught when we carve up the 2d polynomial by the first.
Case i.Is10-ii a factor ofx^3-8?
The rest is 0, hence ten-2 is a gene. In fact,
ten^3-8=(10^ii+2x+4)(10-ii)+0
=(x^2+2x+four)(ten-two)
3.ii Factors Common to All Terms
The method of factoring a common term from a given polynomial is based on the distributive laws, which we recall from Chapter 1,
namely,
ab+ac=a(b+c)
ac+bc=(a+b)c
Example 1. Factor 2x^2+3xy.
The factor common to each term is x. Thus,
2x^2+3xy=ten(2x+3y)
Notice that we can e'er cheque the factorization by multiplying the factors together and comparing the result with the original polynomial.
Case 2.Factor5a^2x^2y^3-25ax^7y^2z.
The mutual factor is5ax^2y^two ,Therefore
5a^2x^2y^3-25ax^7y^2z=5ax^2y^ii(ay-5x^5z)
Case 3.Factor(a+b)x-(a+b)y+(a+b)z.
The common factor is (a + b) and therefore
(a+b)ten-(a+b)y+(a+b)z=(a+b)(10-y+z)
If, in Example 3, the polynomial had been written as
ax+bx-ay-by+az+bz
then no factor is common to all terms. Information technology takes a sure amount of experience to see that the terms tin be grouped so that each grouping has a mutual cistron. For example ax+bx-ay-by+az+bz can be grouped to obtain the expression in Example 2, namely,
ax+bx-ay-past+az+bz=(a+b)ten-(a+b)y+(a+b)z
Sometimes at that place is more than than i fashion to grouping the terms of ii. given polynomial. For example, the above polynomial can be grouped in another way.
ax+bx-ay-by+az+bz=ax-ay+az+bx-by+bz
=a(ten-y+z)+b(x-y+z)
=(a+b)(10-y+z)
Instance 4.Factor20x^2+15xy+8xz+6yz.
20x^2+15xy+8xz+6yz=5x(4x+3y)+2z(4x+3y)
=(5x+2z)(4x+3y)
Example 5.Gene2x^4+4x^3y-2x^2y^3-4xy^4.
Nosotros first observe that 2x is a factor of each term so that
2x^4+4x^3y-2x^2y^3-4xy^iv=2x(x^3+2x^2y-xy^iii-2y^4)
Nosotros at present consider10^three+2x^2y-xy^three-2y^iv. We factorx^ii from first two terms andy^3 from the last ii terms obtaining
x^three+2x^2y-xy^3-2y^four=ten^2(x+2y)-y^3(10+2y)
Each of the two terms has (10 + 2y) equally a factor, so that
x^2(x+2y)-y^three(x+2y)=(10^3-y^three)(x+2y)
Thus,
2x^4+4x^3y-2x^3y^three-4xy^4
=2x(x^3+2x^2y-xy^iii-2y^4)
=2x[x^2(x+2y)-y^iii(x+2y)
=2x[(x^2-y^iii)(10+2y)
=2x(ten^2-y^iii)(ten+2y)
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Example 6.Factor2u^2x-uwx+3uv+6u^2+uvx-3uw.
2u^2x-uwx+3uv+6u^2+uvx-3uw
=u(2ux+wx+3v+6u+vx-3w)
=u[(2ux+6u)+(vx+3v)-(wx+3w)
=u[2u(ten+three)+v(10+3)-due west(ten+3)
=u(2u+v-w)(x+iii)
3.iii The Difference of Two Squares
We observe that
(a-b)(a+b)=a^2-ba+ab-b^2=a^2-b^ii
For purposes of factoring we accept
a^ii-b^two=(a-b)(a+b)
Instance ane.Cistrony^2-ix.
y^ii-9=y^2-3^two=(y-3)(y+3)
Example ii.Factor16x^2-9y^4.
16x^2-9y^iv=(4x)^2-(3y^two)^2= (4x-3y^ii)(4x+3y^2)
Case 3.Factor8u^3z^2-2uz^iv.
8u^3z^2-2uz^4=2uz^2(4u^2-z^2)= 2uz^2(2u-z)(2u+z)
Case four.Factoru^4-16v^4.
u^4-16v^iv
=(u^2)^2-(4v^two)^2
=(u^two-4v^two)(u^2+4v^2)
Sinceu^ii-4v^2is likewise the difference of two squares we accept
u^2-16v^2
=[u^2-(2v)^2](u^2+4v^2)
=(u-2v)(u+2v)(u^ii+4v^2)
Nosotros can as well combine this new idea with the idea of grouping terms to obtain a common factor as is illustrated in the following case.
Example 5.Gene10^2-y^2+4x-4y.
x^2-y^2+4x-4y
=(x^2-y^2)+(4x-4y)
=(10-y)(x+y)+four(x-y)
=(x-y)(x+y+4)
three.iv Polynomials of the Grade 10^2+bx+c
Suppose that x^2+bx+c can be factored into the product of two first caste polynomials; that is,
ten^2+bx+c = (10+r)(x+south)
Since
(x+r)(x+s) =x^two+rx+sx+rs
=x^2+(r+s)x+rs
then
x^2+bx+c =x^2+(r+southward)x+rs
Comparing coefficients we run across that r and due south must satisfy the ii equations
(1)rs=c andr+s=b
Therefore the problem of factoring x^2+bx+c, where b and c are integers, tin can be solved by finding integers r and s that satisfy the 2 equations in (1).
To find integers r and due south that satisfy equations (1), we first find all pairs of integers whose production is c, and so select the pair whose sum is b, if such a pair exists.
Instance one.Factorx^two+5x+6.
The possible factorization of 6 are 6*one and iii*2. In this instance it is easy to encounter that we desire 3*2 since 3+two = v. Thus,
10^ii+5x+half-dozen=(x+two)(x+three)
Case two.Factorten^2-2x-xv.
Hither the factors of -fifteen must be of reverse sign and therefore the possibilities are (15)(- 1), (-15)(1), (v)(-3), and (-5)(3). Since the sum of -5 and 3 is -2, we have
x^2-2x-15 = (ten-5)(x+three)
Instance iii.Factorx^two+2x+3
Since three is prime its only factorization is three*i. Since the sum of these factors is not 2, x^2+2x +three is prime number.
Example 4.Factorx^2-ax-2a^2.
Since the coefficient of x is -a, the only factorization of -2a^2 that we need to consider are (-2a)(a) and (2a)(-a). Observing that -2a+a=-a, we have
x^2-ax-2a^2=(x-2a)(ten+a)
In the next 2 examples we illustrate how this technique may be combined with the ones nosotros take used previously.
Example 5.Factora^2-ab-2b^two+4a+4b.
Since 4a+4b has a factor a+b, we hope that a+b is also a factor of a^2-ab-2b^two. The only factorization of a^2 with a every bit 1 gene is (a)(a) and the only factorization of -2b^2 with b as one factor is (-2b)(b) thus the other factor of a^ii-ab-2b^2 would have to be a-2b. Thus,
a^2-ab-2b^2+4a+4b
=(a^ii-ab-2b^two)+(4a+4b)
=(a+b)(a-2b)+4(a+b)
=(a+b)(a-2b+4)
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Example 6.Factora^four-5a^2+4.
Although this is not a quadratic in a, it is a quadratic in a^2, that is,
a^four-5a^2+4=(a^ii)^2-five(a^2)+4
Since the coefficient of a^2 is negative, the two possible factorization of 4 are (-4) (-1) and (-ii) (-2). Since (-four) + (-1) = -v5, we have
a^iv-5a^2+4=(a^2+4)(a^2-1)
Since each of the factors obtained is the difference of two squares we factor again obtaining
a^4-5a^2+4
=(a^ii-4)(a^2-one)
=(a-2)(a+2)(a-one)(a+1)
3.five Polynomials of the Course ax^2+bx+c
When the coefficient of ten^2 is an integer other than 1, the process of factoring is more difficult. We see by direct multiplication that
(px+q)(rx+s)=prx^2+(ps+qr)x+qs
In comparing this with
ax^2+bx+c
nosotros run into that we must find iv integers p, q, r, and southward such that
pr=a
qs=c
ps+qr=b
Example 1.Cistron2x^2+7x+v.
Suppose 2x^2+7x+5=(px+q)(rx+s). To make certain that nosotros consider all possibilities we conform our piece of work in tabular form. Since all the signs are positive, the only possible factorization of 2 is (two)( 1) and the only possible factorization of v is (5)(1). Nosotros take:
Thus,
2x^2+7x+5
=(px+q)(rx+southward)
=(2x+five)(10+i)
Example two.Factor5y^two+13y-half-dozen.
Suppose 5y^2+13y-six=(py+q)(ry+s). The possible factorizations are tabulated below.
Thus,
5y^2+13y-6=(5y-2)(y+iii)
Instance 3.Cistron8a^2+22ab+15b^2.
Suppose 8a^2+22ab+15b^2=(pa+qb)(ra+sb). The possible factorizations are tabulated below.
Therefore,
8a^ii+22ab+15b^two=(4a+5b)(2a+3b)
Through practise, the pupil will learn to discard certain combinations mentally and this volition cut clown on the number of combinations that he must try.
Certain polynomials provide united states of america with clues as to which combinations we should try.
Instance 4.Factor2x^2+xy-3y^2-8x-12y.
Hither we group the second degree terms and the first degree terms to obtain
2x^2+xy-3y^two-8x-12y
=(2x^ii+xy-3y^2)-four(2x+3y)
In order for the polynomial to factor, we look for 2x^2+xy-3y^ii to accept (2x + 3y) every bit i of its factors. Nosotros obtain
2x^ii+xy-3y^2=(2x+3y)(x-y)
Hence,
2x^2+xy-3y^two-8x-12y
=(2x+3y)(10-y)-4(2x+3y)
=(2x+3y)(ten-y-iv)
Notice that this eliminates the need for a table since there is simply ane possibility that nosotros need endeavour.
3.6 The Sum and Divergence of Two Cubes
We observe that
(a-b)(a^2+ab+b^2)
=a^3+a^2b+ab^2-ba^2-ab^2-b^three
=a^3-b^3
and
(a+b)(a^2-ab+b^2)
=a^3-a^2b+ab^two+ba^2-ab^2+b^iii
a^3+b^3
Thus, a^iii-b^3 and a^3+b^3 can exist factored by the to a higher place equations.
Example 1.Genex^3+27
x^iii+27
=x^3+(3)^3
=(ten+iii)(x^2-3x+9)
Instance 2.Factor2x^eight-16x^2.
2x^8-16x^2
=2x^two(x^vi-8)
=2x^two[(x^two)^iii-two^3
=2x^2(10^2-2)(ten^4+2x^2+4)
Example three.Factor(a-b)^3-(a+b)^iii.
It is sometimes helpful to brand the post-obit type of substitution. Let
A=(a-b) andB=(a+b); and so
(a-b)^iii-(a+b)^iii=A^iii-B^3
=(A-B)(A^2+AB+B^2)
=[(a-b)-(a+b) [(a-b)^2+(a-b)(a+b)+(a+b)^2
=(a-b-a-b) (a^2-2ab+b^ii+a^two-b^two+a^2+2ab+b^2)
=(-2b)(3a^2+b^2)
Case four.Factoru^six-64.
We first annotation that this polynomial can be expressed as either the deviation of two cubes or every bit the difference of two squares, namely,
u^6-64=(u^2)^3-(4)^3
or
u^six-64=(u^iii)^two-(eight)^ii
The computations are simpler if nosotros first gene information technology as the divergence of squares. Therefore,
u^vi-64=(u^3)^2-(8)^2
=(u^iii-viii)(u^three+8)
Each of these factors tin exist factored as the sum or difference of the square. Therefore,
u^6-64
=(u^iii-two^3)(u^three+ii^3)
=(u-2)(u^2+2u+four)(u+two)(u^2-2u+4)
three.7 Polynomials Which Can Be Made the Difference of Two Squares
Sometimes the elementary device of adding and subtracting the square of a monomial can be used to transform a certain type of polynomial, which appears to be nonenforceable, into one that can exist written every bit the difference of two squares.
Outset we find that
(10^2+a)^two=x^4+2ax^2+a^2
Thus, a given polynomial, x^4+bx^2+a^two, will course a perfect square if and only if b = 2a. Now we use this idea to some examples.
Example i.Factorx^iv+x^2+ane
This does not factor by previous methods. Since a=1 we need a coefficient of two for x^2 in order to have a perfect square. Thus nosotros add and subtract x^2 to obtain
x^4+ten^ii+1
=x^4+2x^2+1-10^two
=(x^4+2x^iii+1)-x^ii
=(x^2+1)^2-x^2
The last expression at present factors as the difference of two squares. Therefore,
10^4+ten^2+ane
=(x^2+one)^two-x^2
=[(x^two+1)-x][(x^2+1)+x
=(ten^2-x+ane)(x^2+x+1)
Both of these factors are prime polynomials, and then our factorization is consummate.
Case 2.Cistron9m^4+8m^2+iv
The polynomial does non factor by previous methods. Since 9m^4 and 4 are perfect squares, a middle term of 12m^2 would make the polynomial a perfect square. Thus,
9m^4+8m^two+4
=9m^4+12m^2+4-4m^2
=(3m^2+2)^two-(2m)^2
=[(3m^2+2)-2m][(3m^two+2)+2m
=(3m^ii-2m+two)(3m^two+2m+ii)
Nosotros observe that these two polynomials are prime so that the factorization is complete.
Instance iii.Cistron4x^four-37x^2+9.
Since 4x^4 and 9 are perfect squares, a middle term of -12x^2 would brand the trinomial a perfect square. Therefore,
4x^4-37x^two+9
=(4x^iv-12x^ii+9)-25x^2
=(2x^ii-3)^2-(5x)^2
=(2x^2-5x-3)(2x^2+5x-three)
These polynomials factor, giving the states
4x^iv-37x^2+9
=(2x^2-5x-3)(2x^two+5x-3)
=(2x+i)(ten-three)(2x-1)(x+3)
Find that the original polynomial also factors by the method of Section three.7, which yields
4x^4-37x^2+9
=(4x^2-i)(x^ii-9)
=(2x-i)(2x+ane)(x-3)(10+3)
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2x 2 Xy 3y 2,
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