Question

1. which of the following trinomials can be represented by a rectangle?

(a) ( x^{2}+2 x + 14 )
(b) ( x^{2}+11 x + 30 )
(c) ( x^{2}+10 x + 3 )
(d) ( x^{2}+13 x + 5 )

Explanation:

Step1: Recall the condition for a trinomial to represent a rectangle

A trinomial \(x^{2}+bx + c\) can be represented as a rectangle (factored into \((x + m)(x + n)\) where \(m,n\) are positive integers) if there exist two positive integers \(m\) and \(n\) such that \(m + n=b\) and \(m\times n = c\).

Step2: Analyze option (a) \(x^{2}+2x + 1\)

For \(x^{2}+2x + 1\), we need to find \(m\) and \(n\) such that \(m + n=2\) and \(m\times n=1\). By solving the system \(

$$\begin{cases}m + n=2\\m\times n = 1\end{cases}$$

\), we get \(m = n=1\) since \((x + 1)(x + 1)=x^{2}+2x + 1\).

Step3: Analyze option (b) \(x^{2}+11x+30\)

For \(x^{2}+11x + 30\), we solve \(

$$\begin{cases}m + n=11\\m\times n=30\end{cases}$$

\). We find \(m = 5\) and \(n = 6\) (or vice - versa) because \(5+6 = 11\) and \(5\times6=30\), and \((x + 5)(x + 6)=x^{2}+11x + 30\).

Step4: Analyze option (c) \(x^{2}+10x+3\)

For \(x^{2}+10x + 3\), we try to find positive integers \(m\) and \(n\) such that \(m + n=10\) and \(m\times n = 3\). The factors of \(3\) are \(1\) and \(3\), and \(1+3=4
eq10\). So, it cannot be factored into \((x + m)(x + n)\) with positive - integer \(m\) and \(n\).

Step5: Analyze option (d) \(x^{2}+13x+5\)

For \(x^{2}+13x + 5\), we try to find positive integers \(m\) and \(n\) such that \(m + n=13\) and \(m\times n=5\). The factors of \(5\) are \(1\) and \(5\), and \(1 + 5=6
eq13\). So, it cannot be factored into \((x + m)(x + n)\) with positive - integer \(m\) and \(n\).

Answer:

A. \(x^{2}+2x + 1\), B. \(x^{2}+11x+30\)