Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4
x=5
(3x - y)(3x - 5y) and (2x + 1)(2x + 11)
9x2 - 18x + 99(x2 - 2x + 1)==========
9x2 + 7x = 2 9x2 + 7x - 2 = 0 9x2 + 9x - 2x - 2 = 0 9x(x + 1) - 2(x + 1) = 0 (9x - 2)(x + 1) = 0 x ∈ {-1, 2/9}
9x2+2x-7 = (9x-7)(x+1) when factored
Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4
x=5
(3x - y)(3x - 5y) and (2x + 1)(2x + 11)
(3x2 + 2x + 1)
9x2 - 18x + 99(x2 - 2x + 1)==========
No 9xx-18x+36 9(xx-2x+4) xx-2x+4 (doesn't factor evenly)
9x2 + 7x = 2 9x2 + 7x - 2 = 0 9x2 + 9x - 2x - 2 = 0 9x(x + 1) - 2(x + 1) = 0 (9x - 2)(x + 1) = 0 x ∈ {-1, 2/9}
9x2 = 2x Get the 2x to the left side to get 9x2 - 2x = 0 Use cuadratic formula to solve for x knowing that c is 0 x = [ -b +- sqrt (b2-4ac) ]/2a
-6x + 9x2 - 24 = 3(3x2 - 2x - 8) = 3(3x2 - 6x + 4x - 8) = 3(3x(x - 2) + 4(x - 2)) = 3(3x + 4)(x - 2)
x4 + 2x3 - 9x2 + 18x = x(x3 + 2x2 - 9x + 18) which I do not think can be factorised further.
The answe is 27x^4 + 6x^3 - 61x^2 - 68x - 42