(B) \( t^{2}+4 t-21 \) (C) \( t^{2}+5 t-6 \) (D) \( t^{2}-2 t-8 \) (E) \( t^{2}-10 t-11 \) (F) \( t^{2}+4 t-12 \) (G) \( t^{2}-8 t-20 \) (H) \( t^{2}-t-72 \)

Below are the factorizations of each quadratic expression: (B) t² + 4t – 21   Find two numbers that multiply to –21 and add to 4. Since 7 · (–3) = –21 and 7 + (–3) = 4, we can factor as:     (t + 7)(t – 3) (C) t² + 5t – 6   We need two numbers that multiply to –6 and add to 5. Since 6 · (–1) = –6 and 6 + (–1) = 5, the factorization is:     (t + 6)(t – 1) (D) t² – 2t – 8   Find numbers that multiply to –8 and add to –2. The pair 2 and –4 works because 2 + (–4) = –2 and 2 · (–4) = –8. Thus:     (t + 2)(t – 4) (E) t² – 10t – 11   Look for two numbers that multiply to –11 and add to –10. The numbers –11 and 1 work since (–11) + 1 = –10 and (–11) · 1 = –11. So we have:     (t – 11)(t + 1) (F) t² + 4t – 12   Find numbers that multiply to –12 and add to 4. The numbers 6 and –2 meet the requirement: 6 + (–2) = 4 and 6 · (–2) = –12. Therefore:     (t + 6)(t – 2) (G) t² – 8t – 20   We need two numbers that multiply to –20 and add to –8. The numbers –10 and 2 work because (–10) + 2 = –8 and (–10) · 2 = –20. Factorized, it is:     (t – 10)(t + 2) (H) t² – t – 72   Find two numbers that multiply to –72 and add to –1. The numbers –9 and 8 satisfy these conditions: (–9) + 8 = –1 and (–9) · 8 = –72. Thus:     (t – 9)(t + 8) If you need any further explanation or have another question, feel free to ask!