Given y2 + 5y - 6 = (y + p)(y +q) find the
values of p and q.

p = 6, q = - 1

Step-by-step explanation:

Factor the left side and compare to right side

y² + 5y - 6

Consider the factors of the constant term (- 6) which sum to give the coefficient of the y- term (+ 5)

The factors are + 6 and - 1, since

6 × - 1 = - 6 and 6 - 1 = + 5, thus

y² + 5y - 6 = (y + 6)(y - 1)

Compare with (y + p)(y + q) , then

p = 6 and q = - 1

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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