Factorization problems, factorization problems?

2x 3-x 2+m has a factor of 2x+1, which means that when x=-1 2, 2x 3-x 2+m=0

m=1/2

The most versatile approach.

Polynomials are known to be tripratic.

can be set. 2x+1)*(ax 2+bx+c)=2x 3-x 2+m, then split it into.

2ax^3+(a+2b)x^2+(b+2c)x+c=2x^3-x^2+m

Then let the coefficients correspond one-to-one.

2a=2a+2b=-1

b+2c=0

c=m can be solved.

a=1b=-1

c=-1/2

m=-1/2

You see it's a cubic and the other must have a quadratic, so let's let the other factor be x 2+bx+m

Multiply it by 2x+1.

Then compare the result with the polynomial 2x 3-x 2+m, and the coefficients of the same term are the same, and the help can find m

2x^3-x^2+m=(2x+1)(x^2-x+ 1/2)+ m-1/2

For 2x 3-x 2+m to be factored, there should be m-1 2=0, i.e. m=1 2

x=0, m=1, a common way to fill in the blanks and select, if you are not sure, you can substitute a few more values.

It takes some time to solve the questions, and there are many methods, so ask the teacher, and it is enough to solve this kind of problem.

According to the factoring theorem, when x=, the original formula = 0

So m=1

Polynomials are known to be tripratic.

You can set 2x+1)*(ax 2+bx+c)=2x 3-x 2+m and then split it as.

2ax^3+(a+2b)x^2+(b+2c)x+c=2x^3-x^2+m

Then let the coefficients correspond one-to-one.

2a=2 a+2b=-1

b+2c=0

c=m can be solved.

a=1 b=-1

c=-1/2

m=-1/2

Just use the division of the factor to calculate.

m= Don't understand factoring?

Cross multiplication, two factors split into two adding or subtracting factors, if the result obtained is the same as the original result, the split is correct.

Decompose: 2ax +(a 2) x 1

2ax²+ax－2x－1

2x+1)(ax－1)

2ax +(a 2)x 1 can be broken down into.

2x+1)(ax－1)

That's it.

It is possible to multiply with crosses.

Cross method, go for a while, practice a few more and you can master it.

<> factorization calculus process.

The first 3 terms can be broken down into (x-y), and then 9 can be written as 3 squared, becoming.

x-y)²-3²=0

Then use the squared difference formula, and you get.

x-y-3)(x-y+3)=0

Is it understandable that x-2xy+y-squared is equal to (x-y) squared? Completely flat. Then there's the squared difference formula, which is understandable that 9 equals 3 squared.

When the sum of the coefficients of the odd order terms is equal to the sum of the coefficients of the even order terms, the polynomial has a factor of (x+1);

When the sum of the coefficients of the odd order term is the opposite of the sum of the coefficients of the even order term, the polynomial has a factor of (x-1).

If you understand it from the perspective of equations:

When the sum of the coefficients of the odd term is equal to the sum of the coefficients of the even term, let this polynomial be equal to 0, then x=-1 is one of its roots, for example:

ax 2+bx+c=0, and a+c=b, then substitute x=-1 into ax 2+bx+c=0, which happens to have a+c=b, so:

ax 2+bx+c=0, and a+c=b, then x=-1 is a solution to it.

That is: ax 2+bx+c=0, and a+c=b, then x+1=0 is a condition that satisfies these two equations, thus:

ax 2+bx+c, and a+c=b, then x+1 is a factor of it.

The same is true when the sum of the coefficients of the odd and even terms is inverse.

Factorization, the transformation of a polynomial into the product of several simplest formulas, is called factorization, also known as factorization.

x +2x+1=(x+1) formula method: including square difference, perfect square, cubic difference, sum of cubes, etc.

x +3x+2=(x+1)(x+2) Cross multiplication, also known as differential multiplication, is a widely factored method.

2x+2y =2(x+y) The common factor method, the simplest decomposition method.

Group decomposition method.

Group decomposition is a concise way to solve equations, and let's learn about it.

There are four or more terms of equations that can be grouped and decomposed, and there are two forms of general grouping decomposition: binary division and trione division.

For example: ax+ay+bx+by

a(x+y)+b(x+y)

a+b)(x+y)

We put ax and ay into a group, bx and by into a group, and used the multiplicative distributive property to match the two pairs, which immediately solved the difficulty.

Again, this question can do the same.

ax+ay+bx+by

x(a+b)+y(a+b)

a+b)(x+y)

A few example questions: 1 5ax+5bx+3ay+3by

Solution: =5x(a+b)+3y(a+b).

There are two scenarios for this approach.

Factorization of formulas of type x 2+(p+q)x+pq.

The characteristics of this type of quadratic trinomial are that the coefficient of the quadratic term is 1; The constant term is the product of two numbers; The coefficient of the primary term is the sum of the two factors of the constant term. Therefore, it is possible to factor a quadratic trinomial with a coefficient of 1 for some quadratic terms:

x^2+(p+q)x+pq=(x+p)(x+q) ．

Factorization of formulas of type kx 2+mx+n.

Splitting and adding items.

This method refers to splitting or filling two (or more terms) of a polynomial that are opposite to each other, so that the original formula is suitable for decomposition by the common factor method, the formula method or the grouping decomposition method. Note that the deformation must be performed on the principle of equality with the original polynomial.

For example: BC(B+C)+Ca(C-A)-AB(A+B).

bc(c-a+a+b)+ca(c-a)-ab(a+b)

bc(c-a)+bc(a+b)+ca(c-a)-ab(a+b)

bc(c-a)+ca(c-a)+bc(a+b)-ab(a+b)

bc+ca)(c-a)+(bc-ab)(a+b)

c(c-a)(b+a)+b(a+b)(c-a)

c+b)(c-a)(a+b)．

This is the formula for multiplication. It is also the application of the following formula in the collection:

a+b)(c+d)

a+b)c+(a+b)d.where a, b, c, d are all real numbers.

There's nothing to parse about this simplicity.

You take the underlined part as a whole, multiply it by the two terms in parentheses at the end, and then add it to the right side of the equation.

You can multiply it first and then calculate it. Family.

2ax²+（2a+1）x+1=0

1) When a=0, this is a one-dimensional one-dimensional equation, which is x+1=0, and x=-1(2), when a≠0, this is a one-dimensional quadratic equation.

When it comes to quadratic equations, the cross multiplication method is considered first. 2a 1

2a*1 + 1*1 = 2a +1

So 2ax +(2a+1)x+1=0 can be replaced by (2ax +1）（x+1)=0, i.e. (2ax+1)(x+1)=0

The two roots of this equation are x1=-1 2a, x2=-1

Of course, you can also use the root finding formula, and get the two roots to bring in, which seems to be a required solution, if you use the root finding formula to find it directly.