Question

describe the possible errors made when factoring $x^2 + 14x + 49$ if you get an answer of $(x - 7)^2$.

factor correctly.

Explanation:

To factor \(x^{2}+14x + 49\), we recognize it as a perfect - square trinomial. The formula for a perfect - square trinomial is \(a^{2}+2ab + b^{2}=(a + b)^{2}\). In the trinomial \(x^{2}+14x + 49\), \(a=x\), \(2ab = 14x\), and \(b^{2}=49\). Solving for \(b\) from \(b^{2}=49\), we get \(b = 7\) (since \(2ab=2\times x\times7 = 14x\) which matches the middle term). So, \(x^{2}+14x + 49=(x + 7)^{2}\).

If we get \((x - 7)^{2}\), let's expand \((x - 7)^{2}\) using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\). Here, \(a = x\) and \(b = 7\), so \((x - 7)^{2}=x^{2}-14x + 49\). The original trinomial has a middle term of \(+14x\), while the expansion of \((x - 7)^{2}\) has a middle term of \(- 14x\). The error is in the sign of the middle term when factoring. When we factor a perfect - square trinomial of the form \(a^{2}+2ab + b^{2}\), the factored form is \((a + b)^{2}\), not \((a - b)^{2}\) (unless the middle term is \(-2ab\)). In this case, since the middle term is positive (\(+14x = 2\times x\times7\)), we should have used the plus sign in the binomial square, but instead, a minus sign was used.

Answer:

The error is using a minus sign in the binomial square. The correct factorization of \(x^{2}+14x + 49\) is \((x + 7)^{2}\) because it is a perfect - square trinomial \(a^{2}+2ab + b^{2}\) (with \(a=x\), \(b = 7\)) which factors to \((a + b)^{2}\), while \((x - 7)^{2}=x^{2}-14x + 49\) has a middle term of \(-14x\) that does not match the original trinomial's middle term of \(+14x\). The correct factored form is \(\boldsymbol{(x + 7)^{2}}\).