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One factor of f(x)=5x3-5x2-170x+280is(x+7). What are all the roots of the function? Use the Remainder Theorem.

1 Answer

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Answer:

Roots of the f(x) is -7 , 4 and 2.

Explanation:

Give that

5x^3-5x^2-170x+280 is (x+7)

One factor of f(x) = x + 7

So,

One root , x = -7

First we determine All factors of the given f(x).

Now divide f(x) by given factor the results are given below:

f(x) = ( x + 7 ) ( 5x² - 30x + 40 )

= ( x + 7 ) ( 5x² - 20x - 10x + 40 )

= ( x + 7 ) ( 5x( x - 4 ) - 10( x - 40 ) )

= ( x + 7 ) ( x - 4 ) ( 5x - 10 )

Now

Place x = -7 in the given function,

f(-7) = 5(-7)³ + 5(-7)² - 170(-7) + 280 = -1715 + 245 + 1190 + 280 = 0

So, First root is -7

Now, Place x = 4

f(4) = 5(4)³ + 5(4)² - 170(4) + 280 = 320 + 80 - 680 + 280 = 0

So, Second root is 4

ANd, place x = 10/5 = 2

f(2) = 5(2)³ + 5(2)² - 170(2) + 280 = 40 + 20 - 340 + 280 = 0

So, Third root is 2

Therefore, Roots of the f(x) is -7 , 4 and 2.

User Jitendra Kulkarni
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