Factorization is a subject-predicate phrase,
Factorization is a verb-object phrase,
It is to change polynomials into the form of formula multiplication;
If you need a schematic diagram, just look at the Chinese characters "Mu", "Yue", "You" and "Yong".
"Moon" and "eyes" are two rectangles, 3 in length and a and b in width.
Writing 3a+3b like "Peng" is binomial,
If "moon" and "eye" are spelled into a "use", it is a rectangle of 3(a+b).
The formula of changing 3a+3b into 3(a+b) product is factorization.
Just like decomposing prime factors,
Turn a number into the product of all prime numbers,
To decompose factors, we must thoroughly decompose them.
Minimize the maximum number of each factor,
Specific methods,
The first step is to raise the common factor,
This is also the simplest method.
Common factors include not only coefficients, letters and monomials (we are all familiar with them),
In addition, the common factor can also be a formula,
Like what? (a？ +? b)(3m？ +? 2n)？ +? (2m？ +? 3n)(a？ +? B) What is the common factor? (a+b)
Original style? =? (? Answer? +? b？ )(? 3m？ +? 2n？ +? 2m？ +? 3n？ )?
=? (? Answer? +? b？ )(? 5m？ +? 5n？ )? So we can extract the coefficients again? five
=? 5(? Answer? +? b？ )(? m？ +? n？ )
The second step, formula method,
Is to reverse the formula of algebraic expression multiplication,
A "? -? b "？ =? (a？ -? b)(a？ +? b)？ -Square difference,
A "? +? 2ab？ +? b "？ =? (a？ +? b)"？ -Complete sum of squares,
A "? -? 2ab？ +? b "？ =? (a？ -? b)"？ -Complete square difference,
A ""? +? b " '？ =? (a？ +? b)(a”？ -? ab？ +? b”)？ -Cubes and,
A ""? -? b " '？ =? (a？ -? b )(a”？ +? ab？ +? b”)？ -Cubic difference,
Familiarity with formulas, squares and cubes is the key.
Square difference, and two complete square subtraction formulas,
Like what? 9(? x？ +? y？ )"? -? 4(? x？ +? y？ -? 1? )"
=? [? 3(x？ +? y)？ -? 2(x？ +? y？ -? 1)? ][? 3(x？ +? y)？ +? 2(x？ +? y？ -? 1)? ]
=? (? 3x？ +? 3y？ -? 2x？ -? 2y？ +? 2? )(? 3x？ +? 3y？ +? 2x？ +? 2y？ -? 2? )
=? (? x？ +? y？ +? 2? )(? 5x？ +? 5y？ -? 2? )
Completely flat way, it should be noted that
(? Answer? -? b？ )"
=? [? -? (? b？ -? Answer? )? ]"? =? (? b？ -? Answer? )"
=? A "? -? 2ab？ +? b "？ =? b "？ -? 2ab？ +? One "
and
(a - b )" = [ a + (-b) ]"
= a "？ -? 2ab？ +? b "？ =? A "? +? 2a(-b)？ +? (-b)"
There may be only one formula.
(a？ +? b)"？ =? A "? +? 2ab？ +? b "
Cubic sum, cubic difference and decomposition factor become five terms,
It is difficult to be familiar with the formulas of two linear terms and three quadratic terms.
As long as the specific figures are worked out,
2"'? -? 1? =? 8? -? 1? =? 1? x？ seven
=? (? 2? -? 1? )(? 4? +? 2? +? 1? )
=? (? 2? -? 1? )(? 2"? +? 2? +? 1? )
I just use it? Wheat on the chessboard? Problem, familiar with cubic difference.
A ""? -? 1? =? (? Answer? -? 1? )(? A "? +? Answer? +? 1? ),
A ""? -? b " '？ =? (? Answer? -? b？ )(? A "? +? ab？ +? b？ ),
Cubic difference, the original two cubes are subtracted,
Two linear terms are also subtracted, and all three quadratic terms are added.
Cube and, a "'? +? b " '？ =? (? Answer? +? b？ )(? A "? -? ab？ +? b "？ ),
There is only one quadratic term in the middle? -ab？ It's subtraction, and the rest is addition.
The third step, quadratic trinomial,
I suggest cross multiplication and grouping decomposition together.
Like? x”？ +? (a？ +? b)x？ +? ab？ =? (? x？ +? Answer? )(? x？ +? b？ )
Put the monomial? mx？ =? (a+b)x？ , just open it? Axe? +? bx？ ,
It can be decomposed by common factor grouping.
q？ The key is to look at the positive and negative of the constant term and decide how to divide the principal term into two.
The constant term remains the same, but the first term becomes the opposite number, and the absolute value of the first term divided by two remains the same;
The primary term remains the same, as long as the constant term becomes the opposite number, the primary term will change the way of splitting into two;
q？ If the constant term is a positive number,
One is to disassemble the two items whose absolute values are smaller than the original ones;
Alternatively, the completely flat mode can be decomposed,
x" + 10x + 25
= x" + 5x + 5x + 25
= x( x + 5 ) + 5( x + 5)
= ( x + 5)"
Let's look at x "10x24. There are four factorizations.
x”？ +? 10x？ +? 24
=? x”？ +? 4x？ +? 6x？ +? 24
=? x(？ x？ +? 4? )? +? 6(? x？ +? 4? )
=? (? x？ +? 4? )(? x？ +? 6? )
Constant term? +24? Constant, one-time item? 10x？ Just tear it down, okay 4x？ With what? 6x？ The sum of,
x”？ -? 10x？ +? 24
=? x”？ -? 4x？ -? 6x？ +? 24
=? x(？ x？ -? 4? )? -? 6(? x？ -? 4? )
=? (? x？ -? 4? )(? x？ -? 6? )
q？ If the constant term is negative,
The coefficient of the first term is the difference between the two terms;
x”？ -? 10x？ -? 24
=? x”？ -? 12x？ +? 2x？ -? 24
=? x(？ x？ -? 12? )? +? 2(? x？ -? 12? )
=? (? x？ -? 12? )(? x？ +? 2? )
Constant term? +24? Constant, one-time item? 10x？ Just tear it down, okay 4x？ With what? 6x？ The sum of,
x”？ +? 10x？ -? 24
=? x”？ +? 12x？ -? 2x？ -? 24
=? x(？ x？ +? 12? )? -? 2(? x？ +? 12? )
=? (? x？ +? 12? )(? x？ -? 2? )
Finally, it is necessary to test,
Ensure that the decomposition is thorough and the factorization deformation is correct.
Like what? x^6？ -? Y 6, it should be
=? (? x " '？ -? y’“？ )(? x " '？ +? y " '？ )
=? (? x？ -? y？ )(? x？ +? y？ )(? x”？ -? xy？ +? y "？ )(? x”？ +? xy？ +? y "？ )
Equivalent to? 64? -? 1,
=? (? 8? -? 1? )(? 8? +? 1? )
=? (? 2? -? 1? )(? 4? +? 2? +? 1? )(? 2? +? 1? )(? 4? -? 2? +? 1? )
=? 1? x？ 7? x？ 3? x？ three
If you use the difference three times first, it will work.
=? (? 4? -? 1? )(? 4"? +? 4? +? 1? )
=? (? 2? -? 1? )(? 2? +? 1? )(? 16? +? 4? +? 1? )
=? 1? x？ 3? x？ 2 1
Anything else? 2 1? It is not a prime factor, and the decomposition is incomplete, so it is incorrect.
Just like the square difference now, there are two completely flat subtraction methods.
The formula that needs to be decomposed now is complicated and inconvenient to restore.
We should be familiar with two or three solutions of various formulas.
See if different ways and means have the same result.
In this way, we can check each other and ensure that the answers are correct.