one factor of $f(x)=5x^{3}+5x^{2}-170x + 280$ is $(x + 7)$. what are all the roots of the function? use the…

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.\n$x=-4,x=-2$, or $x = 7$\n$x=-7,x = 2$, or $x = 4$\n$x=-7,x = 5$, or $x = 280$\n$x=-280,x=-5$, or $x = 7$

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.\n$x=-4,x=-2$, or $x = 7$\n$x=-7,x = 2$, or $x = 4$\n$x=-7,x = 5$, or $x = 280$\n$x=-280,x=-5$, or $x = 7$

Answer

Explanation:

Step1: Use polynomial long - division

Since $(x + 7)$ is a factor of $f(x)=5x^{3}+5x^{2}-170x + 280$, divide $5x^{3}+5x^{2}-170x + 280$ by $(x + 7)$ using polynomial long - division or synthetic division. Using synthetic division with $c=-7$: The coefficients are $5,5,-170,280$. Bring down the first coefficient $5$: Multiply $-7\times5=-35$, add to the second coefficient: $5+( - 35)=-30$. Multiply $-7\times(-30) = 210$, add to the third coefficient: $-170 + 210 = 40$. Multiply $-7\times40=-280$, add to the fourth coefficient: $280+( - 280)=0$. The quotient is $5x^{2}-30x + 40$.

Step2: Factor the quotient

Factor out the greatest common factor from $5x^{2}-30x + 40$. The GCF is $5$, so $5x^{2}-30x + 40=5(x^{2}-6x + 8)$. Factor the quadratic $x^{2}-6x + 8=(x - 2)(x - 4)$. So $f(x)=5(x + 7)(x - 2)(x - 4)$.

Step3: Find the roots

Set $f(x)=0$. Then $5(x + 7)(x - 2)(x - 4)=0$. Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. $x+7 = 0$ gives $x=-7$; $x - 2=0$ gives $x = 2$; $x - 4=0$ gives $x = 4$.

Answer:

$x=-7,x = 2$, or $x = 4$