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(4sqrtb-7)^2
\(\bf (7-\sqrt{b})^2\implies (7-\sqrt{b})(7-\sqrt{b})\)
Now what?
wooops missing number there =(
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\(\bf (7-4\sqrt{b})^2\implies (7-4\sqrt{b})(7-4\sqrt{b})\) you'd need to FOIL it
I got 65-56b
hmm is a binomial... so you'd end up with about 3 terms, or a trinomial
16b-56sqrtb+49
or 16b-7sqrtb
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yeap \(\bf (7-4\sqrt{b})^2\implies (7-4\sqrt{b})(7-4\sqrt{b}) \\ \quad \\ 49-28\sqrt{b}-28\sqrt{b}+(4\sqrt{b})^2\implies 49-56\sqrt{b}+16b\)
The one you typed or I typed first?
@jdoe0001
hmmm did you do the FOIL ?
yes
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\(\bf (7-4\sqrt{b})^2\implies (7-4\sqrt{b})(7-4\sqrt{b}) \\ \quad \\ 49-28\sqrt{b}-28\sqrt{b}+(4\sqrt{b})^2\implies \underline{49-56\sqrt{b}+16b}\)
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