(x - a) + (x - a) + (b) = 2 (x - a) + (b) = x - a + x - a + b = 2x - 2a + b
x^2-8x+3 This expression cannot be factored as (x+a)(x+b) where a and b are integers. This is because you cannot find two integers such that their sum is -8 and product 3.
5
11+(7-3)2=
(a x b)^b =ab x b^2 =ab^3
(x - a) + (x - a) + (b) = 2 (x - a) + (b) = x - a + x - a + b = 2x - 2a + b
7.5
Use a^3 + b^3 = (a + b)(a^2 - ab + b^2), where a^2 is a squared, a^3 is a cubed. Note that 216 = 6^3.
3a+ax+3b+bx = 3(a+b)+(a+b)x = (a+b)(3+x)
I am used to going the other way. We will try this. (3,1) = vertex. (X - 3)^2 + 1 = 0 X^2 - 6X + 9 + 1 = 0 X^2 - 6X + 10 = 0 b = -6 c = 10
2x - 13x + 42 = x +ax + b a + b = 2(x - 6.5x + 21) = 34 = a + b
x + x + 1 + x + 2 + x + 3 = -144x + 6 = 144x = 8x = 2
factor b(x+2) + c(x+2) (b+c)(x+2) need more info for futher analysis.
x*2 - 6x + 9 x*2 -3x -3x +9 x(x-3) -3(x-3) (x-3)(x-3) Note also that x*2 - 6x + 9 is of the form a*2 -2ab + b*2 and so the factors are (x-3) and (x-3)
3x2 - 2 is a polynomial of order 2. Therefore, dividing it by (x + 1) will result in a polynomial of order 1. Suppose the quotient is ax + b (where a is non-zero), and with the remainder c. Thus 3x2 - 2 = (x + 1)*(ax + b) + c = ax2 + ax + bx + b + c = ax2 + (a + b)x + (b + c) Comparing coefficients: 3 = a 0 = a + b => 0 = 3 + b => b = -3 -2 = b + c => -2 = -3 + c => c = 1 Therefore, (3x2 - 2)/(x + 1) = 3x - 3 = 3*(x - 1) and a remainder of 1.
Start with the equation 4x2 + 8x + 3 = 0. Invoke the quadratic equation: x = [-b ± √(b2 - 4ac)]/2a, where a = 4, b = 8, and c = 3. Plug in the values for a, b, and c and solve: x = [-8 ± √(82 - 4*4*3)]/2*4, x = [-8 ± √(64 - 48)]/8, x = -1 ± √(16)/8, x = -1 ± 1/2, x = {-1/2, -3/2}.
Factor them. 2 x 2 x b x b = 4b2 2 x 3 x b x b x b = 6b3 Combine the factors, eliminating duplicates. 2 x 2 x 3 x b x b x b = 12b3, the LCM