24-10-2016
Mathematics

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the factors of x²+kx+8 are(x+a)(x+b), where a and b are integers. show that there are two possible values of k

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24-10-2016

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x.x+x.b+x.a+a.b

x^2+(b+a).x+a.b

a.b=8

a=1 b=8

a+b=k

k=1+8=9

a=2 b=4

k=2+4=6

a= -1 b= -8

k= -1-8=-9

a= -2 b= -4

k= -2-4=-6

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