On the polynomial of X 2x ^ 2-11x + m, after factoring the factor, there is a factor x-3. Find the value of M

On the polynomial of X 2x ^ 2-11x + m, after factoring the factor, there is a factor x-3. Find the value of M

On the polynomial of X 2x ^ 2-11x + m, after factoring the factor, there is a factor x-3. Find the value of M
If: 2x ^ 2-11x + M = 0, then x-3 = 0
That is to say, x = 3 is a solution of the above equation
2*9-11*3+M=0
M=15

On the polynomial of X, after the factorization factor 2x? - 11x + m, there is a factor x-4, try to find the value of M

If the factorization factor of the polynomial is 2 (x-4) (x + a), then there is 2x square-11x + M = 2x square + 2 (A-4) x-8a
2(a-4)=-11,m=-8a
A = -- 3 / 2, M = 12

When he solved the equation 2x-1 in 3 is equal to x + A-1 in 2, and then the denominator is removed, both sides are multiplied by 6 at the same time, but the - 1 on the right side of the equation forgets to multiply by 6 Ma Hu's solution to the equation of 2x-1 in 3 is equal to x + A-1 in 2. When the denominator is removed, both sides of the equation are multiplied by 6 at the same time, but the - 1 on the right side of the equation forgets to multiply by 6. Therefore, the solution obtained is x = 4. Try to find the value of a and find the solution of the original equation correctly

(2x-1)/3=(x+a-1)/2
2(2x-1)=3(x+a-1)
-When 1 forgets to multiply by 6, the deformation is as follows:
4x-2=3x+3a-1
x=3a+1=4,
So: a = 1
The correct equation should be:
4x-2=3x+3a-3
X = 3a-1, substituting a = 1 into the calculation:
The solution of the original equation is x = 2

When he solved the equation 2x-1 in 3 is equal to x + A-1 in 2, and then the denominator is removed, both sides are multiplied by 6 at the same time, but the - 1 on the right side of the equation forgets the multiplication

3/2x-1=2/x+a-1
4x-2=3x+3a-1
x=3a+2

When Xiao Ma Hu solved the equation 2x-1 = 2 / 2 x + A-1, when the denominator was removed, the Right-1 of the equation forgot to multiply by 6, so the solution was x = 2. Try to find the value of a and correct the solution Equation

(2x-1) / 3 = (x + a) / 2 - 1, forget to multiply - 1 on the right side of the equation by 6
2(2x-1)=3(x+a)-1
X=2
6=3(2+a)-1
a=1/3
The equation is (2x-1) / 3 = (x + 1 / 3) / 2 - 1
2(2x-1)=3(x+1/3)-6
x=-3

When he solved the equation 2x-1 in 3 is equal to x + A-1 in 2, then the denominator is removed, both sides are multiplied by 6 at the same time, but the - 1 on the right side of the equation forgets to multiply by 6

The - 1 on the right forgot to multiply by 6
It's 4x-2 = 3x + 3a-1
3a=x-1
X=4
So 3A = 4-1 = 3
A=1
Multiply 6 on both sides is 4x-2 = 3x + 3a-6
A=1
So 4x-3x = 3a-4
x=-1

The absolute value of solving equation 3 / 2 times x + 5 is equal to 5

Is it 3 / 2 * / x + 5 / = 5?
/X+5/=5*2/3
/X+5/=10/3
X+5=±10/3
①X+5=10/3 X=-5/3
②X+5=-10/3 X=-25/3

Given the square of ax-3x-18 = (2x-3) (KX + 6), find the values of a and K

Right = (2x-3) (KX + 6) = 2kx ^ 2 + (12-3k) x - 18
Left = ax ^ 2-3x-18
So there are
2k=a （1）
12-3k=-3 （2）
K = 5, a = 10

Ax ^ 2-3x-18 = (2x-3) (KX + 6) find the value of 3a-5k

ax^2-3x-18=2kx^2-(3k-12)x-18
If polynomials are equal, all coefficients must be equal
Therefore, 3K = 3k-3k
K = 5, a = 10
So 3a-5k = 30-25 = 5

If ax ^ 2-3x-18 = (2x-3) (KX + 6), then a=_______ ,k=________ If the product of polynomial X-1 and 2-kx does not contain the first degree of X, then K=_____

ax^2-3x-18
=(2x-3)(kx+6)
=2kx^2+(12-3k)x-18
So a = 2K, 12-3k = - 3
So a = 10, k = 5
(x-1)(2-kx)
=-kx^2+(k+2)x-2
Because it doesn't contain the first term of X
So K + 2 = 0
k=-2