Question

Show that (x - 2) ,(x+3) and (x - 7) are factors of x³- 3x²-10x+24
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Answer 1

x - 2 & x + 3 are factors

Step-by-step explanation:

QUESTION :-

Show that (x - 2) , (x + 3) & (x - 7) are factors of x³ - 3x² - 10x + 24.

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SOLUTION :-

Given polynomial is

p(x) = x³ - 3x² - 10x + 24

To show that,

(x - 2), (x + 3) & (x - 7) are factors of polynomial [p(x)].

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Using Factor theorem*, we will check that these are factors of p(x) or not.

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Let,

x - 2 = 0

=> x = 2

Put this value of x in p(x)

=> p(2) = (2)³ - 3(2)² - 10(2) + 24

= 8 - 3(4) - 20 + 24

= 8 + 24 - 12 - 20

= 32 - 32

= 0

=> p(2) = 0

So, p(x) = 0 for x = 2.

Hence x - 2 is a factor of p(x).

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Let,

x + 3 = 0

=> x = (-3)

Put this value of x in p(x)

=> p(-3) = (-3)³ - 3(-3)² - 10(-3) + 24

= - 27 - 3(9) + 30 + 24

= - 27 - 27 + 30 + 24

= - 54 + 54

= 0

=> p(3) = 0

And, p(x) is 0 for x = - 3.

Hence, x + 3 is also a factor of p(x).

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Now,

Let x -7 be 0

=> x = 7

Put this value of x in p(x).

=> p(7) = 7³ - 3(7)² - 10(7) + 24

= 343 - 3(49) - 70 + 24

= 343 + 24 - 147 - 70

= 367 - 217

= 150

=> p(7) = 150

And, p(x) is 150 for x = 7

Hence, x - 7 is not a factor of p(x).

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So,

x - 2 & x + 3 are factors of x² - 3x² - 10x + 24.

But, x - 7 is not a factor of x³ - 3x² - 10x + 24.

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NOTE :-

Factor Theorem -

If p(a) = 0 for p(x), then x - a is a factor of p(x).

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Hope it helps.

#BeBrainly :-)