Question

determine which of the following polynomial has (x+1) a factor.(1)x4+3x3+3x2+x+1.(2)x3-x2-(2+root2)x+root2​

Answer 1

Answer:

x³ - x² - (2 + √2)x + √2

Step-by-step explanation:

Given the factor is x + 1

so consider

=> x + 1 = 0

=> x = -1.

Now, consider the first polynomial

x⁴ + 3x³ + 3x² + x + 1

substitute x value in this polynomial

=> (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1

=> 1 - 3 + 3 - 1 + 1

=> 1 ≠ 0

X is not a factor of this polynomial.

Lets consider the second polynomial.

x³ - x² - (2 + √2)x + √2

substitute x value in this polynomial also

=> (-1)³ - (-1)² - (2 + √2)(-1) + √2

=> -1 -1 + 2 - √2 + √2

=> 0

so, (x + 1) is a factor of the second polynomial x³ - x² - (2 + √2)x + √2

Answer 2

The zero of x + 1 is -1.

(1) Let p (x) = x4 + 3x3 + 3x2 + x + 1 .

∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1

= 1 – 3 + 3 – 1 + 1 = 1

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1.

(2) Let p (x) = x3 – x2 – (2 + √2) x + √2

∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2

= -1 – 1 + 2 + √2 + √2

= 2√2

⇒ p (-1) ≠ 0

So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2.