Find the cross multiplication formula with coefficient of quadratic term not 1!

Find the cross multiplication formula with coefficient of quadratic term not 1!

I can only say that
Let's start with an example
3 (the square of x) + 6x + 3 = 0 can be written as (3x + 3) (x + 1) = 0
The reason can be seen as follows: the quadratic term can be divided into 3x × x, the primary term is 3x + 3x, and the constant term is 1 × 3. What's the use of this? Why should we divide it in this way? It's all through drawing
3x... 3
x .1
It can be found that vertical multiplication is quadratic term and constant term respectively
If you multiply them and add them together, you will get a term
In writing, according to the symbol, writing horizontally is: (3x + 3) (x + 1) = 0
Because in drafting, there will be cross multiplication, which is the word "ten", so it is called cross multiplication
Let me give you a few more examples to experience
-2X... 3 (decomposition) - 6x + X + 12 = 0, thus (- 2x + 3) (3x + 4) = 0
3x .4
In this problem, we should pay attention to that the first term of multiplication is 9x and - 8x respectively. Do the addition operation to get + X
-2X... - 3 (decomposition) - 6x-17x-12 = 0 is written as (- 2x-3) (3x + 4) = 0
3x .4
It should be noted that the first term of multiplication is - 9x and - 8x respectively. If you do addition, you will get - 17x
And the multiplication of constant terms should be - 12
In the actual operation, you can use this kind of graph to help solve the equation. Of course, it is impossible to see how to split the quadratic term and the primary term and their constant term. As long as you practice more, you can be very skilled and do the problem quickly. The main thing is to master the method!

It is known that X4 + MX3 + nx-16 has factors X-1 and X-2 to find M
4 after X and 3 after X are exponents
How to calculate X4 + MX3 + nx-16 = (x-1) (X-2) (x2 + BX + C) of M and N 1 buildings

Index means the same as you!
Let X4 + MX3 + nx-16 = (x-1) (X-2) (x2 + BX + C) = (x2-3x + 2) (x2 + BX + C) = x4-3x3 + 2x2 + bx3-3bx2 + 2bx + cx2-3cx + 2C = X4 + (B-3) X3 + (2-3b + C) x2 + (2b-3c) x + 2C
Comparing the two sides of the equation, we can get the equations obviously
{
b-3=m
2-3b+c=0
2b-3c=n
2c=-18
b. C is obviously obtainable, and m, n are easy to get the answer
This kind of topic can be calculated like this

If there are two factors x + 1 and X + 2 after the factorization of square + NX + 2 of polynomial MX, then the value of M + n is?

Because: MX + NX + 2 has two factors x + 1 and X + 2, so: x = - 1 and x = - 2 are the two roots of MX + NX + 2 = 0. Substituting x = - 1 and x = - 2 into: M-N + 2 = 0, 4m-2n + 2 = 0, the solution is: M = 1; n = 3, then: M + n = 4

The square of polynomial x plus MX plus 24 can be decomposed into the product of two first-order factors. The value of integer m is -

11. 10, 25, and 14 are all signed
For cross multiplication, suppose M = a + B, a × B = 24, and a and B are integers
From the condition of "a × B = 24, and both a and B are integers", we can get 8 groups of possible values of a and B (we should try them by ourselves to see which two numbers meet the condition), and we can get 8 m by adding them respectively