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If (a+3) divides evenly into ka^3−10a^2+2a+3 , find k
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@Loser66
heeeeeeeeelp please @Loser66
let it =0, replace a = -3 to find k
k=10.3 ?
@Loser66
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well is it??
@Loser66
I got k =-93/27
how?
I got +93/27
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not (-)
ka^3-10a^2+2a+3 =0 ka^3 = 10a^2-2a-3 k = (10a^2-2a-3)/ a^3 replace a = -3 k = (10*(-3)^2-2(-3)-3)/ (-3)^3= 93/ (-27) = -93/27
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yes
ok
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-93/27 = -31/9 Loser66 posted first. Refer to the attachment.
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