If x ^ 2-2x + M can be divided by X + 1, then the value of M is

If x ^ 2-2x + M can be divided by X + 1, then the value of M is

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

What is the value of m if the quadratic trinomial x * 2 + 7x + m of X can be divided by X + 3?
This is a problem of factoring in grade two,

Let x * 2 + 7x + M = k * (x + 3),
When x = - 3, X * 2 + 7x + M = 9-21 + M = 0,
So m = 12

Factorization: X & sup2; (M + n) & sup2; - XY (M + n)
Factorization
x²(m+n)²-xy(m+n)

Extracting common factors
x²(m+n)²-xy(m+n)
=x(m+n)[x(m+n)+y]

A particle jumps from a point one unit from the origin to the origin, and jumps to the midpoint A1 of OA for the first time
For the second time, it jumps from A1 to the midpoint A2 of OA1, and for the third time, it jumps from A2 to the midpoint A3 of oa2. If it keeps jumping, what is the distance from the particle to the origin o after the fifth jump?

The distance from point O after the first jump is 1-1 / 2xoa; the distance from point O after the second jump is 1 - (1 / 2) x (1 / 2) xoa; the distance from point O after the third jump is 1 - (1 / 2) (1 / 2) (1 / 2) xoa The distance between the nth jump and o point is 1 - (1 / 2) of the nth power xoa

How to solve the equation (1.5 + x) x2 = 9
How to solve the equation (1.5 + x) multiplied by 2 = 9

Distribution law by multiplication
1.5×2+x×2=9
2x+3=9
2x=9-3=6
x=6÷2
x=3

Let vector group a1a2a3 be linearly independent, how to prove that a1-a2, A2 = A3, a3-a1 are linearly related

There exists a group of numbers 1,1,1 which are not all zero such that
1(a1-a2)+1(a2-a3) + 1(a3-a1) = 0

An approximate solution (accuracy 0.1) of the equation x & sup2; + 3x-5 = 0 is obtained by dichotomy
The wrong equation is: X & sup3; + 3x-5 = 0

The formula process is: x = (- B ± (square B minus 4ac)) / 2a, the answer is 1.2 or - 4.2

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, if a5a3 = 59, then s9s5 = ()
A. 1B. -1C. 2D. 12

Let the first term of the arithmetic sequence {an} be A1. From the properties of the arithmetic sequence, we can get a1 + A9 = 2a5, a1 + A5 = 2A3, ｜ s9s5 = a1 + a92 × 9a1 + a52 × 5 = 9a55, A3 = 95 × 59 = 1, so we choose a

Given that X1 and X2 are two unequal real roots of the quadratic equation x ^ 2 - (2m + 3) x + m ^ 2 = 0 with respect to x, and satisfying X1 + x2 = m ^ 2, then the value of M is___ .

m=3

As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC is at point D, points E and F are on AB and AC respectively, ∠ BDC = ∠ CDF, indicating the reason why AE = AF

The title should be "BDE = ∠ CDF"
prove:
∵AB＝AC
∴∠B＝∠C
∵AD⊥BC
Ψ BD = CD (isosceles triangle three lines in one: high line, middle line, angle bisector)
∵∠BDE=∠CDF
The value of BDE is equal to CDF
∴BE＝CF
∵AE＝AB-BE,AF＝AC-CF
∴AE＝AF