a. Given a cubic function, \(f(x)=ax^3+bx^2+cx+d\), determine the formulas for the sum and product of the roots of the equation \(f(x)=0\) in terms of \(a\), \(b\), \(c\), and \(d\).

b. If \(p\), \(q\), and \(r\) are the zeros of the function \(f(x) = x^3-7x^2+13x-6\), find the value of

i. \(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{pr}\), and

ii. \((2+p)(2+q)(2+r)\).

a. Given a cubic function, \(f(x)=ax^3+bx^2+cx+d\), determine the formulas for the sum and product of the roots of the equation \(f(x)=0\) in terms of \(a\), \(b\), \(c\), and \(d\).

Solution

a. Let \(r_1\), \(r_2\), and \(r_3\) be the three roots of the equation \(ax^3+bx^2+cx+d=0\).

Then, \(0=\left(x-r_1\right)\left(x-r_2\right)\left(x-r_3\right)\). Expanding and simplifying, we get

Express \(ax^3+bx^2+cx+d=0\) with a leading coefficient of \(1\).

\(x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}=0\), since \(a\neq0\)

a. Given a cubic function, \(f(x)=ax^3+bx^2+cx+d\), determine the formulas for the sum and product of the roots of the equation \(f(x)=0\) in terms of \(a\), \(b\), \(c\), and \(d\).

Solution

\[x^3\textcolor{BrickRed}{+\frac{b}{a}}x^2\textcolor{NavyBlue}{+\frac{c}{a}}x\textcolor{Mulberry}{+\frac{d}{a}}=0\qquad\quad x^3\textcolor{BrickRed}{-\left(r_1+r_2+r_3\right)}x^2\textcolor{NavyBlue}{+\left(r_1r_2+r_2r_3+r_1r_3\right)}x\textcolor{Mulberry}{-r_1r_2r_3}=0\]

Equating the coefficients gives

Therefore, for a cubic equation \(ax^3+bx^2+cx+d=0\), the sum of the roots is given by \(-\frac{b}{a}\) and the product of the roots is given by \(-\frac{d}{a}\).

The sum of the products of all possible distinct pairs of roots is given by \(\frac{c}{a}\).

b. If \(p\), \(q\), and \(r\) are the zeros of the function \(f(x) = x^3-7x^2+13x-6\), find the value of

i. \(\frac{1}{pq}+\frac{1}{qr}+\frac{1}{pr}\).

Solution

i. Simplify \(\dfrac{1}{pq}+\dfrac{1}{qr}+\dfrac{1}{pr}=\dfrac{p+q+r}{pqr}\).

The zeros of the function

are the roots of the equation

so

\(p+q+r\) \(=-\dfrac{b}{a}\) \(=-\dfrac{-7}{1}\) \(=7\)

and

\(pqr\) \(=-\dfrac{d}{a}\) \(=-\dfrac{-6}{1}\) \(=6\)

Hence,

b. If \(p\), \(q\), and \(r\) are the zeros of the function \(f(x) = x^3-7x^2+13x-6\), find the value of

ii. \((2+p)(2+q)(2+r)\).

Solution

ii. Find \((2+p)(2+q)(2+r)\).

If \(p\), \(q\), and \(r\) are the zeros of the function, then \(f(x)=k(x-p)(x-q)(x-r)\). However,

so \(k=1\).

Now,

But, \(f(-2)=(-2)^3-7(-2)^2+13(-2)-6=-68\). Since these are equal,