Factorise 10 x^{2}-x-24 by splitting the midd | vedclass

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(A) To factorise $10 x^{2}-x-24$ by splitting the middle term,we need to find two numbers $p$ and $q$ such that $p+q = -1$ (the coefficient of $x$) an

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$(2 x+3)(5 x-8)$

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(A) To factorise $10 x^{2}-x-24$ by splitting the middle term,we need to find two numbers $p$ and $q$ such that $p+q = -1$ (the coefficient of $x$) and $p 	imes q = 10 	imes (-24) = -240$ (the product of the coefficient of $x^{2}$ and the constant term).We look for factors of $-240$ that add up to $-1$. These numbers are $15$ and $-16$,because $15 + (-16) = -1$ and $15 	imes (-16) = -240$.Now,rewrite the middle term $-x$ as $15x - 16x$:$10 x^{2} - x - 24 = 10 x^{2} + 15 x - 16 x - 24$Group the terms to factor out common factors:$= (10 x^{2} + 15 x) - (16 x + 24)$$= 5 x(2 x + 3) - 8(2 x + 3)$Finally,factor out the common binomial $(2 x + 3)$:$= (2 x + 3)(5 x - 8)$

Factorise $10 x^{2}-x-24$ by splitting the middle term.

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