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8x^{3}+12x^{2}+6x+1-4x^{2}\left(2x+3\right)-7=0
Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(2x+1\right)^{3}.
8x^{3}+12x^{2}+6x+1-4x^{2}\left(2x+3\right)=7
Add 7 to both sides. Anything plus zero gives itself.
8x^{3}+12x^{2}+6x+1-8x^{3}-12x^{2}=7
Use the distributive property to multiply -4x^{2} by 2x+3.
12x^{2}+6x+1-12x^{2}=7
Combine 8x^{3} and -8x^{3} to get 0.
6x+1=7
Combine 12x^{2} and -12x^{2} to get 0.
6x=7-1
Subtract 1 from both sides.
6x=6
Subtract 1 from 7 to get 6.
x=\frac{6}{6}
Divide both sides by 6.
x=1
Divide 6 by 6 to get 1.