Question

x^4+x^3-7x^2-x+6 using factor theorem

Answer 1

$\textbf{ Hello Mate! }$

Thanks for your question!

p(x) = x⁴ + x³ - 7x² - x + 6

Keep once value of x = 1, -1, 2, -2, 3, -3

p(1) = 1⁴ + 1³ - 7(1)² -1 + 6 = 0

p(-1)=(-1)⁴+(-1)³-7(-1)²-(-1)+6=0

p(2) = (2)⁴+(2)³-7(2)²-2+6=0

p(-2) = (-2)⁴ + (-2)³ - 7(-2)² - (-2) + 6 = -12

p(3)=(3)⁴+(3)³-7(3)²-(3)+6=51

p(-3) = (-3)⁴ + (-3)³ - 7(-3)² - (-3) + 6 = 0

Hence, factors are (x+1)(x-1)(x-2)(x+3)

As, degree of p(x) is 4 so there will be 4 factors.

Now, p(x) = k( x + 1 )( x - 1 )( x - 2 )( x + 3 )

x⁴ + x³ - 7x²- x + 6 = k(x+1)(x-1)(x-2)(x+3)

Keep x = 0

We get,

6 = k(1)(-1)(-2)(3)

6 = 6k or k = 1

[tex]\boxed{ Answer:(x+1)(x-1)(x-2)(x+3) }[/tex[

Answer 2

Answer:the factors of y=×⁴+׳-7ײ-2×+6 is y=(×+3)(×+1)(×-1) and roots is -3,2,-1,1

Step-by-step explanation:

its my answer...basta yan mga answer sana makatulong ako