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One factor of One factor of f(x)=5x^3+5x^2-170x + 280 is (x +

7). What are all the roots of the function? Use the Remainder Theorem.

x = –4, x = –2, or x = 7

x = –7, x = 2, or x = 4

x = –7, x = 5, or x = 280

x = –280, x = –5, or x = 7

Verified Answer :

Answer:

Option 2 is correct

Step-by-step explanation:

We have been given an expression

\(5x^3+5x^2-170x+280\)

Since, we have given a factor (x+7) that means at x=-7 the value of given expression will be zero

Similarly we have to check the given points where  they are giving values 0.

Let us check at x=-4 that means put x=-4 in given expression we will get

\(5(-4)^3+5(-4)^2-170(-4)+280\)

We are getting \(f(-4)=720 eq0\)

Hence, x=-4 is dicarded

Hence, option 1 is discarded.

Now, we will check at x=2 we will get

\(5(2)^3+5(2)^2-170(2)+280=0\)

Hence, x=2 is factor of given function.

Now, we will check at x=4

\(5(4)^3+5(4)^2-170(4)+280=0\)

Hence, x=4 is also a factor of given function.

Option 2 is correct

Answer:

x = –7, x = 2, or x = 4 are factors of f(x).

Step-by-step explanation:

Given : f(x)  = 5x³ +5x²-170x +280 and x + 7 is one factor .

To find  : What are all the roots of the function.

Solution :  f(x)  = 5x³ +5x²-170x +280 .

We have given that x + 7 is factor that mean x = -7 is factor of f(x )

When we divide f(x) by x +7 we get

5x² -30x +40 =0

Taking common 5 from equation

5( x² -6x +8) =0

On dividing by 5 both sides

x² -6x +8 = 0

On factoring

x² -2x -4x +8 = 0.

Taking common x from first two terms and -4 from last two terms

x ( x -2) -4 (x -2) = 0

On grouping

(x-2) (x-4) = 0

x -2 =0

x =2

x-4 =0

x = 4

Therefore, x = –7, x = 2, or x = 4 are factors of f(x).

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