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Find all solutions of 7x^4-42x^2-35x=0. I factored it to 7x(x^3-6x-5)=0 but can't get any farther without a graohing calculator.
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7x(x+1)(x^2-x-5)
You can use the fundamental theorem of algebra
x=0 or x=-1then factorise (x^2-x-5)using quadratic equation
(7x(x+1)(x-1)(x-5)
The fundamental theorem of algebra says I'll get 4 solutions, whether real or complex. How does that help me get to 7x(x+1)(x^2-x-5)? That's the step I'm missing.
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