Some students are asked to write down linear and quadratic expressions that have $(x + 2)$ as a factor - Junior Cycle Mathematics - Question 13 - 2014

To determine the value of $k$, we recognize that $x^2 - k$ can be factored as $(x - ext{something})(x + ext{something})$. For $x + 2$ to be a factor, we need to express this as $(x + 2)(x - 2)$, which gives us $k = 4$.

To find Denise's expression, we calculate:

$(x + 2)(2x + 3) = x(2x + 3) + 2(2x + 3) = 2x^2 + 3x + 4x + 6 = 2x^2 + 7x + 6$

Thus, Denise's expression is $2x^2 + 7x + 6$.

Division allows determining if $x + 2$ is a factor by checking if the polynomial $3x^2 + 11x + 10$ divides evenly. If the result has no remainder, then $x + 2$ is a factor.

Performing polynomial long division:

1. Divide the leading term: 3x^2 ig/x = 3x.
2. Multiply: $3x(x + 2) = 3x^2 + 6x$.
3. Subtract: $(3x^2 + 11x + 10) - (3x^2 + 6x) = 5x + 10$.
4. Divide again: 5x ig/x = 5.
5. Multiply: $5(x + 2) = 5x + 10$.
6. Subtract: $(5x + 10) - (5x + 10) = 0$.

The remainder is $0$, thus $(x + 2)$ is a factor. The quotient is therefore $3x + 5$.

One example of a quadratic expression that has $(x + 2)$ as a factor is:

$(x + 2)(x + 3) = x^2 + 5x + 6.$

This shows that $(x + 2)$ is a factor of the quadratic expression.