One factor of f(c) = 5x3 + 5x2 - 170x + 280 is (x + 7). what are all the roots of the function? use the remainder theorem

So the roots of the polynomial are -7 , 4 , 2

Step-by-step explanation:

Given the polynomial in the question

5x³ + 5x² - 170x + 280

One of the factor of this polynomial is (x + 7)

By using remainder theorem

          5x²-30x+40

x+7 | 5x³ + 5x² - 170x + 280

        5x³ + 35x²

                  -30x² -170x + 280

                  -30x² -210x

                              40x + 280

                              40x + 280

                                             0

So the polynomial we have is  

5x²- 30x + 40

divide by 5

x² - 6x + 8

(x-4)(x-2)

So the roots of the polynomial are -7 , 4 , 2

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

Step-by-step explanation:

Given:

One factor of f(x) = x + 7

⇒ One root , x = -7

To find: All roots of the function Using Remainder theorem.

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

On dividing f(x) by given factor we get,

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 )

Remainder Theorem: Remainder theorem states that if a polynomial p(x) is divided by a linear polynomial of form x -a then remainder is given by p(a).

Put 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, Put x = 4

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

So, Second root is 4

Now put 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.

in the attachment.   i didn't see your choices.   the word which means there is choices.

step-by-step explanation:

(x,y)

x tells you how much left or right you go from the origin

y tells you how much down or up you go from the origin

so for example (-3,2), you go left 3 and up twice. i denoted which of the ordered pairs is which in my graph.