Let p=x^2-7p=x 2 −7p, equals, x, squared, minus, 7. Which equation is equivalent to (x^2-7)^2-4x^2+28=5(x 2 −7) 2 −4x 2 +28=5left parenthesis, x, squared, minus, 7, right parenthesis, squared, minus, 4, x, squared, plus, 28, equals, 5 in terms of ppp ? Choose 1 Choose 1 (Choice A) A p^2+4p-5=0p 2 +4p−5=0p, squared, plus, 4, p, minus, 5, equals, 0 (Choice B) B p^2-4p+23=0p 2 −4p+23=0p, squared, minus, 4, p, plus, 23, equals, 0 (Choice C) C p^2-4p-5=0p 2 −4p−5=0p, squared, minus, 4, p, minus, 5, equals, 0 (Choice D) D p^2+4p+23=0p 2 +4p+23=0 Help Plz

4 puzzles in 3 hours

step-by-step explanation:

cheater

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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