How to solve factorization

How to solve factorization

The formula should be clear in your mind, which formula can be applied when you see the problem. The only way is to do more exercises to make the formula memory
principle:
1. The factorization must be thorough (that is, the factorization after factorization cannot be done again)
2. Only parentheses were left at the end of the result
3. The polynomial first term of the result is positive. Extract the common factor from a formula
Where, is the common factor. Therefore, the answer after factorization is: formula reorganization
Reorganize the formula and then extract the common factor
Cross multiplication
Main entry: cross multiplication
Sum of two squares or difference between two squares (see square difference)
Based on
principle:
1. The factorization must be thorough (that is, the factorization after factorization cannot be done again)
2. Only parentheses were left at the end of the result
3. The polynomial first term of the result is positive. Extract the common factor from a formula
Where, is the common factor. Therefore, the answer after factorization is: formula reorganization
Reorganize the formula and then extract the common factor
Cross multiplication
Main entry: cross multiplication
Sum of two squares or difference between two squares (see square difference)
According to the above two identities, if the original formula meets the above conditions, it can be directly decomposed by using the pronoun. For example, it can be decomposed into.
Sum and difference of two n-th power numbers
Sum of two cubes
Can be decomposed into
The difference between two cubes
Can be decomposed into
The difference between two n-th power numbers
The sum of two odd power numbers
Familiar with the formula can be solved, on the several formulas
First of all, we should observe that if we use the formula properly, we should use the formula. If we don't use the formula, we should make up the formula
A question about factorization in grade two
X-Y = 1, xy = 3, find the value of X & sup3; y-2x & sup2; Y & sup2; + XY & sup3~
x³y-2x²y²+xy³
=xy(x^2 -2xy+y^2)
=xy(x-y)^2
=3×1^2
=3
x^3y-2x^2y^2+xy^3
=xy(x^2-2xy+y^2)
=xy(x-y)^2
=3*1^2
=3
The original formula = XY (X & sup2; - 2XY + Y & sup2;) = XY (X-Y) 2 = 3
From a to B is an uphill road, from B to C is a downhill road, a student from a to B to C, immediately return to a along the original road, share 3.5 hours. Known uphill speed is the same, downhill speed is also figured out, and it takes 30 minutes more to walk uphill than downhill, how long does it take students to walk uphill?
Uphill x hours, downhill y hours
x+y=3.5
x-y=0.5
X=2
y=1.5
Let T1 be the uphill time and T2 be the downhill time.
Then the formula {T1 + T2 = 3.5, T1-T2 = 0.5}
So T1 = 2 hours, that is, 2 hours for uphill road
The uphill time is x and the downhill time is y
x-Y=30
x+y=3.5*60
x=120 y=90
A mathematical problem about set in grade one of senior high school
Given the equation AX + B = 0, when a and B satisfy what conditions, the solution set of the equation is finite? When a and B satisfy what conditions, the solution set of the equation is infinite?
Move the term first, ax = - B,
1. When a is not equal to 0, x = - B / A, then the equation is a finite set
That is, a is not equal to 0, B is any real number
2. When a = 0, it can't be further solved, then B = 0, then x is any real number and satisfies 0 = 0. Therefore, when AB is all 0, the equation is an infinite set
I wish you a happy study~
Ask a mathematical problem of quadratic equation of two variables
A two digit number minus 5 times the sum of its two digits, the result is 15. The quotient of this two digit number divided by the sum of its two digits is 6, and the remainder is 3?
Suppose: ten digit is α and one digit is β
（10α＋β）－（α＋β）×5＝15
（10α＋β）÷（α＋β）—3=6
Connect these two equations to form a system of equations and then solve them
I wish you better and better results
The topic is...
If there is only one element a in the set a = [x I xx + ax + B = x], find the values of a and B
Because a is an element of a set, x = a is the solution of XX + ax + B = x equation, that is 2A & sup2; - A + B = 0
In addition, the set a has only one element a, so the equation 2A & sup2; - A + B = 0 has two equal real roots, that is △ = 1-8b = 0. ｜ B = 1 / 8, a = 1 / 4
Because there's only one, so I put a in
a^2+a^2+b=a
2a^2-a+b=0
Because there's only one a
So 1-4 * 2 * b = 0
So B = 1 / 8, substituting a = 1 / 4
Mr. Wang has 30000 yuan in his hand and wants to buy the three-year treasury bonds with an annual interest rate of 2.89%. When he arrives at the bank, the remaining treasury bonds are less than 30000 yuan. After Mr. Wang buys all these treasury bonds, the remaining money is deposited in a three-year fixed-term bank deposit with an annual interest rate of 2.7%, and he has to pay 20% interest tax when due How many yuan of treasury bills did you buy at the end of the year and how much yuan did you deposit in the bank?
Let's say X Yuan for treasury bills and Y yuan for bank deposit. According to the meaning, we can get x + y = 30000 30000 + 3 × 2.89% x + 3 × 2.7% Y (1 − 20%) = 32338.2, and the solution is x = 18000y = 12000. A: Mr. Wang bought 18000 yuan treasury bills and deposited 12000 yuan in the bank
Ask a math problem of senior one, it should be about complement: u = R, a = {x | x2}, B = {y | y = x square + a}, if CUA is included in B, find the value range of real number a
Because we know the set a, CUA is greater than or equal to - 1 and less than or equal to 2, set B can be changed to y greater than or equal to a, because CUA is a subset of B, so the value range of a is less than or equal to - 1
Hand fight
Cherish!
Mathematical problems of quadratic equation of two variables
There is a mother and daughter. Five years ago, the mother's age is 15 times that of her daughter. 15 years later, the mother's age is more than twice that of her daughter. 6. What's the age of the mother and daughter now?
Daughter is 7 years old, mother is 35 years old, 15 (X-5) = y-52 (x + 15) + 6 = y + 15
Let the age of the daughter be X. Mother for y
Then: 15 (X-5) = Y-5
2(X+15)+6=15+Y
The solution is x = 7. Y = 35
Mother 35
Daughter 7
Let the present age of the mother be a and the present age of the daughter be B
a-5=15*（b-5）
a+15=2*（b+15)+6
A = 35, B = 7
So the mother and daughter are now 35 and 7, respectively
Let the mother's age be x and the daughter's age be y
X-5=15(Y-5)
X+15=2（Y+15）+6
Solve this system of equations
Y=7 X=35
Finding a ∩ B from the set a {y | y = x & # 178; - 1, X ∈ r}, B {y | y = - 2x & # 178; + 2, X ∈ r}
A = {y | y = x & # 178; - 1, X ∈ r} denotes the range of function y = x & # 178; - 1
Namely
　　 A=[-1,+∞)
B = {y | y = - 2x & # 178; + 2, X ∈ r} denotes the range of function y = - 2x & # 178; + 2
Namely
B=(-∞,2]
so
　　A∩B=[-1,+∞)∩(-∞,2]=[-1,2]