PLEASE HELP: Find all the real zeros of this polynomial function?

y=x^4-x^3-6x^2

y= x²(x²-x-6)

y= x²(x-3)(x+2)

y = 0 for x = 0, x= -2, x = 3 ............answer

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utilising the rational root theorem, attempt all distinctive factors of the *final* term (-6) divided via all distinctive factors of the 1st coefficient (a million, subsequently, that's implied in front of the x3). So, you attempt -6/a million , -3/a million , -2/a million , -a million/a million , a million/a million , 2/a million , 3/a million , and six/a million or -6, -3, -2, -a million, a million, 2, 3, and six The a million's are easiest to purpose, So, permit's attempt plugging in +a million, first... f(a million) = a million + 2 - 5 - 6 = -8 (does not artwork) attempt -a million... f(-a million) = -a million + 2 + 5 - 6 = 0 (works!) because of the fact x=-a million works, then we would desire to continuously be waiting to ingredient out a (x - -a million)... or, in different words, a (x + a million). Doing long branch of polynomials, we get (x3 + 2x2 - 5x - 6) / (x + a million) = (x2 + x - 6) So, f(x) = (x + a million)(x2 + x - 6) can we ingredient that 2nd polynomial? properly, if the 2nd polynomial would properly be factored into something like (x + a)(x + b), all of us understand, from FOIL, that a*b will would desire to return out to be -6 and that a+b will would desire to be the coefficient of the 'x' term, that's +a million. What can 'a' and 'b' be such that a*b=-6 and a+b=a million ? properly, that is a=3 and b=-2, suited? So, we would desire to continuously be waiting to ingredient the 2nd polynomial into (x + 3)(x - 2). And, lo, if we divide (x2 + x - 6) via (x + 3), we get (x - 2). So, it works. So, thoroughly factored, f(x) = (x + a million)(x + 3)(x - 2) and the zeros would be the values of x which reason *any* of those binomials to be 0... that are x = {-a million, -3, +2} For you, that is answer D