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Question
Using the Factor Theorem, show that (x + 5) is a factor of 2x3 + 5x2 – 28x – 15. Hence, factorise the expression 2x3 + 5x2 – 28x – 15 completely.
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Solution
Let f(x) = 2x3 + 5x2 – 28x – 15
x + 5 = 0 `\implies` x = –5
∴ Remainder = f(–5)
= 2(–5)3 + 5(–5)2 – 28(–5) – 15
= –250 + 125 + 140 – 15
= –265 + 265
= 0
Hence, (x + 5) is a factor of f(x).
Now, we have,
2x2 – 5x – 3
`x + 5")"overline(2x^3 + 5x^2 - 28x - 15)`
2x3 + 10x2
– –
– 5x2 – 28x
– 5x2 – 25x
+ +
– 3x – 15
– 3x – 15
+ +
0
∴ 2x3 + 5x2 – 28x – 15 = (x + 5)(2x2 – 5x – 3)
= (x + 5)[2x2 – 6x + x – 3]
= (x + 5)[2x(x – 3) + 1(x – 3)]
= (x + 5)(2x + 1)(x – 3)
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