Question

factor the following trinomial. 7c² - 4c - 20 (7c + ?)(c - )

Explanation:

Step1: Let the missing numbers be \(a\) and \(b\), so the factored form is \((7c + a)(c - b)\). Expand this: \(7c^2 - 7bc + ac - ab = 7c^2 + (a - 7b)c - ab\).

Step2: Compare with \(7c^2 - 4c - 20\). So we have two equations: \(a - 7b = -4\) and \(ab = 20\).

Step3: Solve \(ab = 20\) for possible integer pairs \((a,b)\): (1,20),(2,10),(4,5),(-1,-20),(-2,-10),(-4,-5). Test in \(a - 7b = -4\).

For \(a = 10\), \(b = 2\): \(10 - 7\times2 = 10 - 14 = -4\), which works.

Answer:

The first box is \(10\) and the second box is \(2\), so the factored form is \((7c + 10)(c - 2)\).