Question

Show that (x+4),(x-3) and (x-7) are factors of x³-6x²-19x+84.

Answer 1

Answer:

We showed that  (x+4),(x-3) and (x-7)  are the factors for the given polynomial $x^3-6x^2-19x+84$

Step-by-step explanation:

Given polynomial $x^3-6x^2-19x+84$

Let (x+4),(x-3) and (x-7) be the factors for the given  polynomial.

To show that  (x+4),(x-3) and (x-7) are factors for the given polynomial :

By using the Synthetic Division we can solve this cubic expression.

First take x-7 is a factor

7_|  1     -6     -19      84

       0     7       7      -84

      _________________

       1      1      -12      0

• Hence the given cubic equation satisfies with the factor x-7
• Therefore x-7 is a factor
• Now we have the quadratic equation
• $x^2+x-12=0$
• $(x-3)(x+4)=0$

• x-3 and x+4 is also factors for the given polynomial

• Therefore the factors are  (x+4),(x-3) and (x-7)
• The given polynomial can be written as  $x^3-6x^2-19x+84=(x+4)(x-3)(x-7)$

Therefore (x+4),(x-3) and (x-7)  are the factors for the given polynomial $x^3-6x^2-19x+84$

Hence showed.