Factor.
-$t^{2}+1.4 t+0.49$
A) $\left(t^{2}+0.7\right)^{2}$
B) $(t+0.7)^{2}$
C) $(t+1.4)^{2}$
D) $t(t+1.4)$

Name: Factor. -$t^{2}+1.4 t+0.49$ A) $\left(t^{2}+0.7\right)^{2}$ B) $(t+0.7)^{2}$ C) $(t+1.4)^{2}$ D) $t(t+1.4)$
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Question

Quizplus · Accepted Answer

This is a perfect square trinomial. Factoring it gives $(t+0.7)^{2}$ because $0.7 \times 0.7 = 0.49$ and $2 \times 0.7 = 1.4$.

Factor - t2+1.4t+0.49t^{2}+1.4 t+0.49t2+1.4t+0.49 A) (t2+0.7)2\left(t^{2}+0.7\right)^{2}(t2+0.7)2 B) (t+0.7)2(t+0.7)^{2}(t+0.7)2 C) (t+1.4)2(t+1.4)^{2}(t+1.4)2 D) t(t+1.4)t(t+1.4)t(t+1.4)

Factor
- $t^{2}+1.4 t+0.49$ A) $\left(t^{2}+0.7\right)^{2}$ B) $(t+0.7)^{2}$ C) $(t+1.4)^{2}$ D) $t(t+1.4)$