Question

Plz solve this question for me it's the 1st question plz it's 12 at night and I am figuring how to do this plz I beg of u

Answer 1

recall that for any root say x - a, namely x - a = 0, thus x = a, for such a root of a polynomial, f(x), based on the remainder theorem f(a) = 0.

to put it in different lingo, since we know that we have two factors (x - 2 ) and (x+1), that simply means we have two roots of x = 2 and x = -1, well, using the remainder theorem that simply means f(2) = 0 and f(-1) = 0.

$\bf x^3+ax^2+2x+b\\\\ -------------------------------\\\\ \stackrel{x=2}{f(2)}=(2)^3+a(2)^2+2(2)+b\implies \stackrel{\stackrel{f(2)}{\downarrow }}{0}=8+4a+4+b \\\\\\ \boxed{0=4a+b+12} \\\\\\ \stackrel{x=-1}{f(-1)}=(-1)^3+a(-1)^2+2(-1)+b\implies \stackrel{\stackrel{f(x)}{\downarrow }}{0}=-1+a-2+b \\\\\\ 0=a+b-3\implies \boxed{3-b=a}\\\\ -------------------------------$

$\bf \stackrel{\textit{\textit{using the a value in the first equation found}}}{0=4(3-b)+b+12}\implies 0=12-4b+b+12 \\\\\\ 0=24-3b\implies 3b=24\implies b=\cfrac{24}{3}\implies b=8\\\\ -------------------------------\\\\ \stackrel{\textit{and since we know that }}{3-b=a}\implies 3-8=a\implies -5=a\\\\ -------------------------------\\\\ x^3-5x^2+2x+8$

and you can graph that, you'll see the roots or x-intercepts of -1 and 2 there.