Evaluate b^2 – 2bc + c^2 if b = -4 and c = –2
I am lost on this one. Any help would be fabulous. Thanks in advance.
You first plug in -4 for b and -2 for c.
(-4)^2 - 2 (-4)(-2) + (-2)^2
Then, following order of operations (PEMDAS), solve it.
16 - 2 (8) + 4
16 - 16 + 4
0 + 4
4
So the answer is 4.
(-4)^2 - 2(-2*-4) + (-2)^2
16-(2*8)+4
16-16+4= 4
(-4)^2 - 2(-4*-2) + (-2)^2 = 4
(-4)^2 = 16
2(-4*-2) = 16
(-2)^2 = 4
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Verified answer
You first plug in -4 for b and -2 for c.
(-4)^2 - 2 (-4)(-2) + (-2)^2
Then, following order of operations (PEMDAS), solve it.
16 - 2 (8) + 4
16 - 16 + 4
0 + 4
4
So the answer is 4.
(-4)^2 - 2(-2*-4) + (-2)^2
16-(2*8)+4
16-16+4= 4
(-4)^2 - 2(-4*-2) + (-2)^2 = 4
(-4)^2 = 16
2(-4*-2) = 16
(-2)^2 = 4