Chapter 8.5, Problem 41PPE

a)

To determine

To find: Information about p and q when $c>0$ , in equation $a{x}^2+bx+c=(x+p)(x+q)$ .

Answer to Problem 41PPE

Values of both p and q will be positive or negative. Or, when $c>0$ , then either $p>0,q>0$ or $p<0,q<0$ .

Explanation of Solution

Given information: Identity $a{x}^2+bx+c=(x+p)(x+q)$ is given.

Formula used: As $a{x}^2+bx+c$ is quadratic polynomial its two factors are $(x+p)(x+q)$ .

Calculation: On simplifying above equation,

  $\begin{array}{l}a{x}^2+bx+c=(x+p)(x+q)\\ =x(x+q)+p(x+q)\\ =x^2+qx+px+pq\\ =x^2+(p+q)x+pq\end{array}$

On comparing constant terms on each side,

  $c=pq$

And c is positive only if both p and q are positive or both should be negative as,

  $(i)(-)\times (-)=(+)(ii)(+)\times (+)=(+)$

Conclusion: So, for being c positive, the values of p and q both should either be positive or negative.

b)

To determine

To find: Information about p and q when $c<0$ , in equation $a{x}^2+bx+c=(x+p)(x+q)$ .

Answer to Problem 41PPE

Values of either of p and q will be negative. Or, when $c<0$ , then either, $(i)p>0,q<0$ or $(i)p<0,q>0$.

Explanation of Solution

Given information: Identity $a{x}^2+bx+c=(x+p)(x+q)$ is given.

Formula used: As $a{x}^2+bx+c$ is quadratic polynomial its two factors are $(x+p)(x+q)$ .

Calculation: On simplifying above equation,

  $\begin{array}{l}a{x}^2+bx+c=(x+p)(x+q)\\ =x(x+q)+p(x+q)\\ =x^2+qx+px+pq\\ =x^2+(p+q)x+pq\end{array}$

On comparing constant terms on each side,

  $c=pq$

And c is negative only if either of p or q should be negative because, $(-)\times (+)=(-)$ .

Conclusion: So, for being c negative, either of p or q should be negative.