Let f (x) = the third power of x minus the second power of x plus KX plus 4, K be a constant. It is known that f (x) ≡ (X-2) (the second power of AX + BX + C), a, B and C are constants a) Find a, B, C b) Someone thinks that all the roots of F (x) = 0 are real numbers,

Let f (x) = the third power of x minus the second power of x plus KX plus 4, K be a constant. It is known that f (x) ≡ (X-2) (the second power of AX + BX + C), a, B and C are constants a) Find a, B, C b) Someone thinks that all the roots of F (x) = 0 are real numbers,

f(x)=x³-x²+kx+4;
f(x)=(x-2)(ax²+bx+c)=ax³+bx²+cx-2ax²-2bx-2c=ax³+(b-2a)x²+(c-2b)x-2c;
So a = 1; b-2a = - 1; c-2b = k; - 2C = 4;
∴a=1;b=1;c=-2;k=-4;
(2)f(x)=(x-2)(x²+x-2)=(x-2)(x+2)(x-1);
So it's right. It's all real numbers
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Given the quadratic function y = x square + BX + C, the image passes through points (1,3), (4,0)
Given the quadratic function y = x square + BX + C, the image passes through points (1,3), (4,0)
Finding the value of B and C

By substituting the coordinates of known points into the analytical formula, the solution can be obtained by solving the binary linear equations about B and C;
The image of quadratic function y = x & # 178; + BX + C passes through points (1,3), (4,0),

It is known that the parabola y = a times the square of X + BX + C (a > o) and the straight line y = K (x-1) - quarter K square, no matter any real number is taken, there is only one parabola and straight line
There is only one common point between the parabola and the straight line
There are many symbols that can't be expressed

ax^2+bx+c=kx-k-0.25k^2 ax^2+(b-k)x+c+k+0.25k^2=0 （a>