If \(\dfrac{1}{x+k}+\dfrac{1}{x+2k}+\dfrac{1}{x+5k}=\dfrac{1}{k}\), then which one of the following is the solution of this equation?
\(k\)
Substitute \(x = k\): \(\dfrac{1}{2k}+\dfrac{1}{3k}+\dfrac{1}{6k} = \dfrac{3+2+1}{6k} = \dfrac{6}{6k} = \dfrac{1}{k}\), which satisfies the equation. (Setting \(x=tk\) and simplifying gives \(t^3+5t^2+t-7=0\), i.e. \((t-1)(t^2+6t+7)=0\); the other two roots are irrational and negative, so \(t=1\), i.e. \(x=k\), is the matching solution.)
If (x 2- 1) is a factor of ax 4+ bx 3+ cx 2+ dx + e, then which one of the following is correct?
If α, β and γ are the zeros of the polynomial f(x) = ax 3+ bx 2+ cx + d, then α 2+ β 2+ γ 2is equal to
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The HCF and the LCM of two polynomials are 3x + 1 and 30x 3 + 7x 2 - 10x - 3 respectively. If one polynomial is 6x 2 + 5x + 1, then what is the other polynomial?
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What should be added to \(\frac{1}{{(x - 2)(x - 4)}}\) to get \(\frac{2x - 5}{{(x^2-5x+6)(x - 4)}}\) ?
What is (x - a) (x - b) (x - c) equal to?
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Factorize x 2- y 2- 9z 2+ 6yz
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