In order to factorize the quadratic trinomial X & # 178; - 5x + P in the range of integers, how many values can the integer P have?

In order to factorize the quadratic trinomial X & # 178; - 5x + P in the range of integers, how many values can the integer P have?

Let x2-5x + P
＝(x+a)(x+b)
＝x^2+(a+b)x+ab
therefore
a+b=-5
ab=p
Because there is no array of integers satisfying a + B = - 5
So there are countless values of P
For example: take a = 1, B = - 6, then p = - 6
(if it is a positive integer, then there are only finite groups)
25/4
6X & # 178; + 7x = - 6, please x =? Thank you for your help!
6x²+7x+6=0
Δ=7^2-4*6*6
X equals 6 out of 43
First, judge △ = b * b-4ac, and you can know △
Solve two equations: (1) {3x + 2Y = - 2, X-Y = - 4 (2) {3 / X-2 / y = 1, 3x-4y = 11, quick!
1）
3x+2y=-2,(1)
x-y=-4(2)
3x-3y=-12(3)
(1)-(3)
5y=10
Y=2
x-y=-4
x=-2
Y=2
（2）
1/3x-1/2y=1,(1)
3x-4y=11 (2)
3x-9/2y=9(3)
(2)-(3)
1/2y=2
Y=4
3x-4y=11
X=9
Y=4
1 x= -2,y= 2; 2 x=5 ,y= 1
In order to factorize the quadratic trinomial xx-5x + P in the range of integers, how many values do P have
Why?
It can be x ^ - 5x - 4, it can be x ^ - 5x, it can be x ^ - 5x + 4, it can be x ^ - 5x + 6.25, so p = - 40 4 and 6.25
Know x2 = Y3, find the value of 7X2 − y2x2 − 2XY + 3y2?
Let x = 2K, y = 3k, the original formula = 7 · (2k) 2 − (3K) 2 (2k) 2 − 2 · 2K · 3K + 3 · (3K) 2 = 19k219k2 = 1
If the solution of the system of equations x-2y = 3, the solution of 3x + 4Y + 5 is x = 2.2, y = - 0.4, then the system of equations
（a+2009)-2(b-2010)=3 3(a+2009)+4(b-2010)=5
Finding the value of a and B
a+2009=2.2
b-2010=-0.4
a=-2006.8
b=2009.6
x=a+2009
y=b-2010)
x-2y=3
3x+4y=5
10y=2
y=1/5
b=2010+1/5
Another similar!
Compare the above formula x = a + 2009 y = b-2010 a = - 2006.8 B = 2009.6
Another x = a + 2009;
y=b-2010;
From this we know that x = 2
y=-0.4;
A = 2007
b＝2009.6;
In order to factorize the quadratic trinomial x2 + PX + 5 in the range of integers, how many values of P are there
X2 + PX + 5 can be factorized in the range of integers ｛ X & ｛ 178; + PX + 5 = 0x1 × x2 = 5, X1 + x2 = - P5 = 1 × 5 = - 1 × - 5 ｛ - P = 1 + 5 = 6, P = - 6 or - P = - 1-5 = - 6, P = 6 ｛ P = ± 6. This is my conclusion after meditation. If it can help you, I hope you will give me a satisfactory reply
The quadratic trinomial x2 + PX + 5 can be factorized in the range of integers
5=1*5=(-1)*(-5)
So there is p = 1 + 5 = 6 or - 1-5 = - 6
There are two kinds
Given X & # 178; + Y & # 178; = - 2XY, find the value of X-Y / x + 3Y
x²+y²+2xy=0
（x+y）²=0
X + y so x = - y
So X-Y / x + 3Y = - Y-Y / - y + 3Y
=-2y/2y
=-1
You can ask me if you don't understand
x²+y²=-2xy
x²+y²+2xy=0
(x+y)²=0
x=-y
x-y=x+x=2x
x+3y=x-3x=-2x
(x-y)/(x+3y)=2x/(-2x)=-1
(x+y)²=0，x=-y
(x-y)/(x+3y)=-2y/(2y)=-1
Because X & # 178; + Y & # 178; = - 2XY, x + y = 0, x = - Y.
（X-Y）/（X+3Y）=2X/-2Y=-1
x²+y²=-2xy
x²+y²+2xy=0
(x+y)²=0
x+y=0
x=-y
x-y／x+3y
=(-y-y)/(-y+3y)
=-1
To solve the linear equation of three variables: x + 4Y + 3Z = 8,4x + y + 2Z = 7,2x + 2y-7z = - 3
3y-16z
=2(2x+2y-7z)-(4x+y+2z)
=(-3)*2-7
=-13
15y+10z
=4(x+4y+3z)-(4x+y+2z)
=4*8-7
=25
According to 3y-16z = - 13
15y+10z=25
So, 90z = (15y + 10z) - 5 * (3y-16z) = 25-5 * (- 13) = 90
So z = 1, y = 1, x = 1
Factorization 1. π R ^ 2 - π R ^ 2 2. X ^ 3-4x ^ 2 + 4x 3. - 4 (x-2y) ^ 2 + 9 (x + y) ^ 24. A ^ 3c-4a ^ 2BC + 4AB ^ 2C
①πR^2-πr^2 ②x^3-4x^2+4x ③-4(x-2y)^2+9(x+y)^2 ④a^3c-4a^2bc+4ab^2c
πR^2-πr^2=π（R^2-r^2）=π（R+r）(R-r)x^3-4x^2+4x=x(x^2-4x+4)=x(x-2)^2-4(x-2y)^2+9(x+y)^2 =9(x+y)^2 -4(x-2y)^2=(3x+3y)^2-(2x-4y)^2=(5x-y)(x+7y)a^3c-4a^2bc+4ab^2c=ac(a^2-4ab+4b^2)=ac(a-2b)^2