If (x-1) is a factor of x power 3-7x power 2+14x-8
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If (x – 1) is a factor of x3 – 7x2 + 14x – 8 then on putting x – 1 = 0
x = 1
f(1) = 0
= 13 – 7(1)2 + 14(1) - 8
= 1 – 7 + 14 – 8 = 0
Hence x – 1 is one factor.
To find other factors = x3 – 7x2 + 14x – 8
= x2(x – 1) – 6x(x – 1) + 8(x – 1)
= (x – 1)(x2 – 6x + 8)
= (x – 1)(x2 – 4x – 2x + 8)
= (x – 1){x(x – 4) – 2(x – 4)}
= (x – 1)(x – 2)(x – 4).
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