Question

show that (X + 4,) (x-3)(x-7) are the factors of xcube -6xsquare- 19x+84​

Answer 1

Step-by-step explanation:

for

${x}^{3} - 6 {x}^{2 } - 19x + 84$

for showing that x+4, x-3, x-7 are solutions of above equation

we have to put value of x in equation

before that we have to fiind value of x

we find that by putting them equal to zero

first part x+4=0

x=-4

(-4)*(-4)*(-4)-6(-4)(-4)-19(-4)+84

-64-96+76+84

=0

same you can do with others

Answer 2

SOLUTION:-

Let f(x)= x³ -6x² -19x +84

By factor theorem, we know that,

If p(x) is a polynomial and a is any real number,then g(x)= (x-a) is a factor of p(x), if p(a)=0

For checking (x+4) to be a factor, we will find f(-4)

=) f(-4)= (-4)³ -6(-4)² -19(-4)+84

=) f(-4)= -64 - 96 +76 +84

=) f(-4)= -160 + 160

=) f(-4)= 0

So, (x+4) is a factor.

For checking (x-3) to be a factor, we will find f(3)

=) f(3)= (3)³ - 6(3)² -19(3) +84

=) f(3)= 27 - 54 -57 + 84

=) f(3)= -27 + 27

=) f(3)= 0

So, (x-3) is a factor.

For checking (x-7) to be a factor, we will find f(7)

=) f(7)= (7)³ -6(7)² -19(7) +84

=) f(7) = 343 -294 - 133 + 84

=) f(7)= 49 -49

=) f(7)= 0

So, (x-7) is a factor.

Therefore,

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

Hope it helps ☺️