How to factorize the general quadratic trinomial ax ^ 2 + BX + C

How to factorize the general quadratic trinomial ax ^ 2 + BX + C

In general, for a quadratic trinomial, if:
The equation AX ^ 2 + BX + C = 0 can only be decomposed if it has real roots, otherwise it cannot be decomposed
That is, when a, B and C satisfy B ^ 2-4ac > = 0, the equation AX ^ 2 + BX + C = 0 has real roots X1 and X2 (which may be equal), and the polynomial can be decomposed into:
ax^2+bx+c=(x-x1)(x-x2)
If we can't use cross multiplication or undetermined coefficient method, we have to use formula method.
{x+[b+√(b^2-4ac)]/(2a)}{x+[b-√(b^2-4ac)]/(2a)}
Y = x & # 178; - 6x + 7, X belongs to [0, a], find the range of this function,
How to classify the process or idea
f(x)=x²-6x+7
=(x-3)²-2
The axis of symmetry is x = 3
So the cut-off points are 3 and 6
（1）0
Given the equations 3x + 2Y = 14 x-3y = 2, the following deformation is correct: a 12x + 8y = 4 12x-9y = 6 d 9x + 6y = 3 8x-6y = 4
Which one is right!
Both are right
If the quadratic trinomial X & sup2; - AX-8, (a is an integer), it can be factorized in the range of integers
-8=-1*8=-2*4=-4*2=-8*1
So - a = - 1 + 8
-a=-2+4
-a=-4+2
-a=-8+1
So a = - 7, a = - 2, a = 2, a = 7
The original formula = the square of X - (the square of Y + 2Y + 1)
=The square of X - (y + 1)
=(x+y+1)(x-y-1)
It's a perfect square
Consider a + B as X and 6 as y
Then x ^ 2-2xy + y ^ 2 = (a-y) ^ 2 = (a + B-6) ^ 2
What is the range of the function y = Log1 / 2 (X & # 178; - 6x + 17)?
Why is this (x-3) &# 178; + 8 ≥ 8
(x-3)²+8＞0
From the system of equations 5x-y = 110, 9y-x = 110, we can get the equation of degree one variable as
From 9y-x = 110, we get - x = 110-9y, x = 9y-110
So 5 * (9y-110) - y = 110
45y-550-y=110
44y=660
y=15
Substituting x = 9y-110
x=9*15-110= 25
In order to factorize the quadratic trinomial X & # 178; - mx-6 in the integer range, the integer values of M can be taken as ± 5 and ± 1
Multiplication by cross
6 can only be decomposed into 1 * 6 or 2 * 3
So m can only be 6-1,1-6,2-3 or 3-2
That is ± 5 and ± 1
If the explanation is not clear enough,
In fact, it is obtained by decomposing and adding the constant terms
It should be wrong... It can't be broken down. Are you sure it's right???
First of all, we need to make the form of (x-a) (X-B) = 0, and after the expansion, the constant term is ab, we can get AB = 6. Because the two integers can only be 2 * 3 or 1 * 6, so the values of a and B are 2 or 3 and 1 or 6 respectively, and then we can get the four values of m by substituting the values into each other
Find the range of function y = (x ^ 2 + 7x + 10) / (x + 1) (x is not equal to - 1)
As soon as possible
y=（x^2+7x+10）/（x+1）
=(x+2)(x+5)/（x+1）
x+1=t
y=(t+1)(t+4)/t
=(t^2+5t+4)/t
=t+4/t+5
When t is greater than 0
≥4+5=9
When t is less than 0
≤-4+5=1
The range is negative infinity to 1 and 9 to positive infinity
BZD
x^2+7x+10=yx+y
X ^ 2 + (7-y) x + 10-y = 0 has roots (obviously not - 1)
So △ = (7-y) ^ 2-4 (10-y) > = 0
y^2-10y+9>=0
y> = 9 or Y
-7x / 3Y ^ 2 times (- 9y ^ 2 / x ^ 2) calculation process,
-7x -9y^2
-----×（___________ )
3y^2 x^2
-7x and x ^ 2 become - 7 and X
3Y ^ 2 and - 9y ^ 2 become 1 and - 3
The original formula = - 7 × (- 3) / X
=21/x
If the quadratic trinomial X & # 178; - ax + 15 is factorized in the range of integers, then the integer a is acceptable? (you need to fill in an answer that you think is correct.)
-8 = - 1 * 8 = - 2 * 4 = - 4 * 2 = - 8 * 1, so - a = - 1 + 8-A = - 2 + 4-A = - 4 + 2-A = - 8 + 1, so a = - 7, a = - 2, a = 2, a = 7