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Question

Factorise: 2x310x24x+24
using synthetic division.

A
2(x+2)(x3)(x4)
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B
(x+2)(x3)(x4)
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C
2(x2)(x3)(x4)
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D
2(x+2)(x3)(x+4)
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Solution

The correct option is A 2(x+2)(x3)(x4)
p(x)=(2x310x24x+48)
Using trial and error method,
when
x=2p(2)=2×(2)310×(2)24×(2)+24) =16+40848=0
Hence, one factor = (x+2)

Using synthetic division,
x3x2x1x021044824284821424|0

p(x)=(x2)(2x214x+24)

Now, using trial and error method,
when,
x=3q(x)=(2x214x+24) =2×3214×3+24=0
Hence, another factor is (x3)

x2x1x021424362428|0

p(x)=(x+2)(x3)(2x8) =2(x+2)(x3)(x4)

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