rove that x^3+4x +7 has no integral zeros
its x grade question(age 13 -14)without using calculus
Using the rational roots theorem
Rational root theorem - Wikipedia, the free encyclopedia
The only possible rational and hence possible integral zero are
$\displaystyle \pm 1 \pm 7$
and since none of these are zero's we are done
Hello, ayushdadhwal!
$\displaystyle \text{Prove that }\,x^3+4x +7\,\text{has no integral zeros. }$
$\displaystyle \text{It's x grade question (age 13 -14) without using calculus.}$
$\displaystyle \text{Let }\,y \:=\:x^3+4x+7\,\text{ and plot some points on its graph.}$
. . $\displaystyle \begin{array}{c||c|c|c|c|c|} x & \text{-}3 & \text{-}2 & \text{-}1 & 0 & 1 \\ \hline y & \text{-}32 & \text{-}9 & 2 & 7 & 12 \end{array}$
We find that the graph is strictly increasing
. . and the only x-intercept occurs between (-2, -9) and (-1, 2)
. . (when the graph crosses the x-axis).
Hence, the only zero of the function occurs between $\displaystyle x = \text{-}1$ and $\displaystyle x = \text{-}2.$
Therefore, there are no integral zeros.