Question

1. error analysis describe and correct the error a student made in making a table in order to factor the trinomial $x^2 - 11x - 26$. \

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factors	sum of factors	\
---	---	\
$-1$ and $11$	$10$	\
$1$ and $-11$	$-10$	\

\
the trinomial $x^2 - 11x - 26$ is not factorable because no factors of $b$ sum to $c$.

Explanation:

Step1: Recall factoring trinomial \(x^2 + bx + c\)

For a trinomial \(x^2+bx + c\), we need two numbers \(m\) and \(n\) such that \(m\times n=c\) and \(m + n=b\). For \(x^2-11x - 26\), \(c=- 26\) and \(b=-11\). So we need two numbers \(m\) and \(n\) where \(m\times n=-26\) and \(m + n=-11\).

Step2: Analyze the student's error

The student was looking for factors of \(b\) (coefficient of \(x\)) instead of factors of \(c\) (constant term). Also, the student should consider factors of \(- 26\) (since \(c=-26\)) not just factors related to \(11\). The correct approach is to find two numbers that multiply to \(-26\) and add up to \(-11\). Let's find such numbers: we need \(m\times n=-26\) and \(m + n=-11\). The numbers are \(-13\) and \(2\) because \((-13)\times2=-26\) and \(-13 + 2=-11\). So the trinomial factors as \((x - 13)(x+2)\). The student's error was in the set of factors considered (wrongly focused on factors related to \(b\) instead of \(c\)) and concluded it's not factorable, but it is factorable.

Answer:

The student's error was: 1. Did not consider factors of the constant term \(-26\) (instead considered factors related to the coefficient of \(x\), \(-11\)). 2. Incorrectly concluded the trinomial is not factorable. The correct factoring: find two numbers \(m\) and \(n\) with \(m\times n=-26\) and \(m + n=-11\). The numbers are \(-13\) and \(2\) (since \((-13)\times2=-26\) and \(-13 + 2=-11\)). So \(x^2-11x - 26=(x - 13)(x + 2)\)