To find the zeros of the function it is necessary and sufficient to solve the equation : The zeros of the function will be the roots of this equation. Thus, the zeros of the function are at the point .

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x^{2}-x-6=0-x+3\gt 2x+1; line\:(1,\:2),\:(3,\:1) f(x)=x^3; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos

For zeros, we first need to find the factors of the function x^{2}+x-6. The factors of x^{2}+x-6are (x+3) and (x-2). Now we equate these factors with zero and find x. i.e., x+3=0and

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The Quadratic formula is a formula for finding the zeros of a quadratic function. Let ax^ {2} + bx +c = 0 ax2 + bx + c = 0 be a quadratic function where a, b, c are constants with a

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