Question

what mistake was made when factoring the trinomial below? $6x^2 + 39x + 60$ $6x^2 + 39x + 60$ gcf: 3 $3(2x^2 + 13x + 20)$ $40x^2$ $+8x$ $+5x$ $+13x$ $40x^2$ $2x$ $+4$ $2x^2$ $+8x$ $+5$ $+5x$ $+20$ answer: $(2x + 5)(x + 4)$

Explanation:

Step1: Check GCF factoring

The original trinomial is \(6x^2 + 39x + 60\). The GCF of 6, 39, and 60: \( \gcd(6,39,60) = 3\), so factoring out 3 gives \(3(2x^2 + 13x + 20)\), which is correct so far.

Step2: Check factoring \(2x^2 + 13x + 20\)

To factor \(ax^2+bx+c\) (here \(a = 2\), \(b = 13\), \(c = 20\)), we need two numbers that multiply to \(a\times c=2\times20 = 40\) and add to \(b = 13\). The numbers 8 and 5: \(8\times5 = 40\), \(8 + 5 = 13\). So we rewrite the middle term: \(2x^2+8x + 5x+20\).

Step3: Group and factor

Group as \((2x^2 + 8x)+(5x + 20)\). Factor out GCF from each group: \(2x(x + 4)+5(x + 4)\). Then factor out \((x + 4)\): \((2x + 5)(x + 4)\). Wait, but wait—wait, when we expand \((2x + 5)(x + 4)\), we get \(2x^2+8x+5x + 20=2x^2+13x + 20\), which is correct. But wait, the original trinomial after factoring out 3 is \(3(2x^2 + 13x + 20)\), and then factoring \(2x^2 + 13x + 20\) gives \((2x + 5)(x + 4)\), so the full factorization should be \(3(2x + 5)(x + 4)\). Wait, but the mistake—wait, no, wait the problem is about the mistake in the given factoring. Wait, no, wait the given answer in the image is \((2x + 5)(x + 4)\) without the GCF 3. Oh! The mistake is that after factoring out the GCF (3) from the original trinomial, when factoring the quadratic, they forgot to include the GCF 3 in the final answer. The correct factorization should be \(3(2x + 5)(x + 4)\), but the given answer is \((2x + 5)(x + 4)\), missing the GCF 3 that was factored out initially.

Answer:

The mistake was forgetting to include the greatest common factor (GCF) of 3 in the final factored form. The correct factorization should be \(3(2x + 5)(x + 4)\), but the given answer omitted the 3 that was factored out initially.