Factorization: 12x ^ 5-7x ^ 4 + 19x ^ 3-20x ^ 2 + 11x-15

Factorization: 12x ^ 5-7x ^ 4 + 19x ^ 3-20x ^ 2 + 11x-15

12x5-7x4+19x3-20x2+11x-15
=12(x5-x4)+5(x4-x3)+24(x3-x2)+4(x2-x)+15(x-1)
=(x-1)(12x4+5x3+24x2+4x+15)
12x4 > | 5x3 | 24x2 > | 4x in 12x4 + 5x3 + 24x2 + 4x + 15|
The original formula can no longer have a factor of (x + a)
So the formula can only have the factor (x2 + ax + b) (x2 + CX + D)
15 = 5 * 3, 5A + 3C = 4, the most likely is 5 * (- 1) + 3 * 3 = 4
12=3*4
3 * 3 + (- 1) * 4 = 5
So 12x4 + 5x3 + 24x2 + 4x + 15 = (3x2-x + 3) (4x2 + 3x + 5)
3 -1 3
4 3 5
Double crosses multiply
12X^5-7X^4+19X^3-20X^2+11X-15
=(x-1)(3x2-x+3)(4x2+3x+5)
7x ^ 4 + 20x ^ 3 + 11x ^ 2 + 40-6 factorization
the sooner the better
7x^4+20x^3+11x^2+40x-6
=7x^4+21x^3-x^3-3x^2+14x^2+42x-2x-6
=7x^3(x+3)-x^2(x+3)+14x(x+3)-2(x+3)
=(x+3)(7x^3-x^2+14x-2)
=(x+3)[x^2(7x-1)+2(7x-1)]
=(x+3)(7x-1)(x^2+2)
A
A
A
How to solve the equation when 5x + 3 (56-x) = 216
5X+3（56-X）=216
5X+168-3X=216
2X=216-168
X=48÷2
X=24
If M is a positive real number and 1 of M-M = 3, then M & # 178; + M & # 178; 1 of M = 3
m-1/m=3
Square on both sides
m²-2+1/m²=9
m²+1/m²=11
The equation AX2 - (3a + 1) x + 2 (a + 1) = 0 has two unequal real roots X1 and X2, and x1-x1x2 + x2 = 1-A, then the value of a is ()
A. 1b. - 1C. 1 or - 1D. 2
According to the meaning △ 0, namely (3a + 1) 2-8a (a + 1) > 0, namely a2-2a + 1 > 0, (A-1) 2 > 0, a ≠ 1, ∵ the equation AX2 - (3a + 1) x + 2 (a + 1) = 0 of X has two unequal real roots X1 and X2, and x1-x1x2 + x2 = 1-A, ∵ x1-x1x2 + x2 = 1-A, ∵ X1 + x2-x1x2 = 1-A, ∵ 3A + 1a-2
Solve the equation 1 / 20x + 56-2 / 5x = 42 x =?
1/20x+56-2/5x=42
can scarcely wait
can 't wait
1/20x+56-2/5x=42
56+-7/20X=42
-7/20x=-14
x=40
It is known that the real number x satisfies that x + X / 1 is equal to 3, and X & # 8308; + 2mx & # 178; + 1 / 2 x & # 179; + MX & # 178; + X is equal to 1 / 3?
From x + 1 / x = 3 ①, we get (for the square of both sides): X & # 178; + 1 / X & # 178; = 9-2 = 7 ②
The numerator and denominator of the original fraction are divided by X & # 178
(x+m+1/x)/(x²+2m+1/x²)=1/3
(3 + m) / (7 + 2m) = 1 / 3
So m = 2
Thank you for your adoption!
The equation AX2 - (3a + 1) x + 2 (a + 1) = 0 has two unequal real roots X1 and X2, and x1-x1x2 + x2 = 1-A, then the value of a is ()
A. 1b. - 1C. 1 or - 1D. 2
According to the meaning △ 0, namely (3a + 1) 2-8a (a + 1) > 0, namely a2-2a + 1 > 0, (A-1) 2 > 0, a ≠ 1, ∵ the equation AX2 - (3a + 1) x + 2 (a + 1) = 0 of X has two unequal real roots X1 and X2, and x1-x1x2 + x2 = 1-A, ∵ x1-x1x2 + x2 = 1-A, ∵ X1 + x2-x1x2 = 1-A, ∵ 3A + 1a-2
(x + 56) / 3 = (X-38) / 2 to solve the equation
Every step of the way
(x+56)/3=(x-38)/2
2(x+56)=3(x-38)
x=226
2（x+56）=3（x-38）
2x+112=3x-114
2x-3x=-114-112
-x=-226
x=226
So the solution of the original equation is x = 226 (x + 56) / 3 = (X-38) / 22 (x + 56) = 3 (x -... Expansion)
2（x+56）=3（x-38）
2x+112=3x-114
2x-3x=-114-112
-x=-226
x=226
So the solution of the original equation is x = 226, the answer is: (x + 56) / 3 = (X-38) / 22 (x + 56) = 3 (X-38) 2x + 112 = 3x-114 2x-3x = - 114-112 - x = - 226 x = 226, so the solution of the original equation is x = 226