Factorization: (a ^ 2 + 3a) ^ 2 - (A-1) ^ 2

Factorization: (a ^ 2 + 3a) ^ 2 - (A-1) ^ 2

=(a²+3a-a+1)(a²+3a+a-1)
=(a²+2a+1)(a²+4a-1)
=(a+1)²(a²+4a-1)

（x2+3x+12）(x2+7x+12)-120
The first polynomial is 12! Not 2!

If it's 12, there's no way to break it down. Sorry
（X^2+3x+2）(x^2+7x+12)-120
=(x+1)(x+2)(x+3)(x+4)-120
=(X^2+5x+4）(x^2+5x+6)-120
=(X^2+5x+4）^2+2(X^2+5x+4）-120
=(X^2+5x+4+12)(X^2+5x+4-10)
=(X^2+5x+16）(X^2+5x-6）
=(x-1)(x+6)(X^2+5x+16）

(2a + B · b) - 4A · a factorization!

If a-2a + 1 = 0, then 2a-4a + 5=

∵a²-2a+1=0
∴a²-2a=-1
∴2a²-4a+5=2(a²-2a)+5=2×（-1）+5=3

How to draw a square in Geometer's Sketchpad?

The method of drawing a square in the Geometer's Sketchpad: 1. One is to draw a line segment first, and then construct a vertical line through the two ends of the line segment, and then make two circles with the line segment as the radius and the two ends as the center of the circle. The two circles have two intersections with the vertical line in front, which can be connected. 2. The other is to use the "rotation" in the "transformation" to first

When k takes what value, the univariate quadratic inequality 2kx Λ 2 + kx-3 / 8 < 0 holds for all real numbers x?

Combined with the image of quadratic function of one variable
2kx^2+kx-3/8

As shown in the figure, point P is a point in parallelogram ABCD, s triangle ABP = 7, s triangle AOP = 4, find s triangle ACP

11

Given the quadratic function y = = - x2 + X-1 / 5, when the independent variable is M-1, the corresponding function value is greater than o, when the independent variable x is M-1 respectively
Given the quadratic function y = = - x2 + X-1 / 5, when the independent variable is M-1, the corresponding function value is greater than o, when the independent variable x is M-1, M + 1, the corresponding function value is Y1, Y2, then Y1, Y2 meet the requirements

Let - x2 + X-1 / 5 = 0, the relationship between the two is 0

Z = sin (x, y) finding the partial derivative of Z to X at (1,1)

Sin (x, do it as sin (x × y)!
Z′x＝ycos（xy）.Z′x|（1,1）＝cosi
Z′y＝xycos（xy）.Z′y|（1,1）＝cosi

Put two different pieces in the 5 * 5 grid, or how many kinds of square hair are there in the same row and not in the same column

Let's put one piece first, 9 squares in the horizontal and vertical rows of this chess piece, no other piece, 25-9 = 16 kinds (the number of the first chess piece put at one time) 16 × 25 = 400 kinds (the first chess piece can put 25 positions × the front 16 kinds)