Question

x^+3x-40 factorise by splitting middle term
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Answer 1

$\bf\blue{\underline{\boxed{Question}}}$

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$\small\tt{x^{2} + 3x - 40\: =\: 0}$

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$\bf\blue{\underline{\boxed{Solution}}}$

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$\small\tt{x^{2} +8x - 5x - 40\: =\: 0}$

$\small\tt{x^{2} + 8x -5x - 40}$

$\small\tt{x(x+8) -5(x+8)}$

$\small\tt{(x-5)\:(x+8)}$

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$\small\tt\red{\underline{\boxed{x\:-5\:=\:0\:\rightarrow\:5}}}$

$\small\tt\red{\underline{\boxed{x\:+8\:=\:0\:\rightarrow\:-8}}}$

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$\bf\blue{\underline{\boxed{Thanks}}}$

Answer 2

Answer:

the equation is : x^2+3x-40

Step-by-step explanation:

so,

ax^2+bx-c

a=1, b=3, c= -40

a*c= ( 1 )( -40 )

a*c= -40

now,

find the factors of -40 whose sum or difference is +3.

factors = 5 and 8

= 5*8 = 40

now, splitting the middle term

=> x^2 +5x - 8x - 40

(change the signs of the factors to fit the condition)

=> x(x+5) -8(x+5)

=> (x+5) (x-8)

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