If x^{2}-10x+21=(x+m)(x+n),then m+n = ldots | vedclass

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(D) Given the quadratic expression $x^{2}-10x+21 = (x+m)(x+n)$.Expanding the right side,we get $(x+m)(x+n) = x^{2} + nx + mx + mn = x^{2} + (m+n)x + m

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$-10$

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(D) Given the quadratic expression $x^{2}-10x+21 = (x+m)(x+n)$.Expanding the right side,we get $(x+m)(x+n) = x^{2} + nx + mx + mn = x^{2} + (m+n)x + mn$.Comparing the coefficients of $x$ and the constant term with the original expression $x^{2}-10x+21$,we get:$m+n = -10$$mn = 21$Therefore,the value of $m+n$ is $-10$.

If $x^{2}-10x+21=(x+m)(x+n)$,then $m+n = \ldots$

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