Question

1. Determine which of the following polynomial has (x+1) a factor
a)x power 4 +x power 3 +x power 2 +x+1


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Answer 1

Correct Question

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Check whether the x+1 is factor of

x⁴ + x³ + x² + x + 1 or not

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The above Question falls from Polynomials

Solution

Let P(x) = x⁴ + x³ + x² + x + 1

F(x) = x + 1

F(x) = 0

➱ x + 1 = 0

➱ x = -1

By Remainder Theorem, (x + 1) will be a factor of P(x), if P(x) = 0

Now,

P(-1) = x⁴ + x³ + x² + x + 1

⠀⠀⠀➦ (-1)⁴ + (-1)³ + (-1)² + (-1) + 1

⠀ ⠀⠀➦ 1 + (-1) + 1 + (-1) + 1

⠀ ⠀⠀➦ 1 - 1 + 1 - 1 + 1

⠀ ⠀⠀➦ 3 - 2

⠀ ⠀⠀➦ 1

Here, we got F(x) = 1

The remainder is not equal to zero hence (x+1) is not a factor of the given polynomial.

Answer 2

A polynomial if has a number as it's factor ; the simplifications and value of this polynomial ; gives 0 .

To find of the given polynomial is a factor ; first we need to find the value of g(x) and then substituting the same in p(x) and at last try to get zero .

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Step 1 : Find value of g(x)

For finding the value of g(x) we need to make the given factor in form of zero equation :

$\small{\longrightarrow{\sf{g(x) = x + 1 = 0 }}}$

If we transpose +1 to the other side its sign changes to -1 .

$\small{\longrightarrow{\sf{g(x) = x = 0 - 1 }}}$

$\small{\longrightarrow{\sf{g(x) = x = -1 }}}$

So the value of g(x) is (-1)

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Step 2 : Put the value of x in polynomial:

$\small{\longrightarrow{\sf{p(x) = {(x)}^{4} + {(x)}^{3} + {(x)}^{2} + x + 1}}}$

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Step 3 : Simplification

$\small{\longrightarrow{\sf{p(-1) = {(-1)}^{4} + {(-1)}^{3} + {(-1)}^{2} + (-1) + 1}}}$

$\small{\longrightarrow{\sf{p(-1) = 1 + (-1) + 1 + (-1) + 1}}}$

$\small{\longrightarrow{\sf{p(-1) =\cancel{ 1 + (-1)} \cancel{+ 1 + (-1)} + 1}}}$

$\large{\boxed{\implies{\sf{p(-1) = 1 }}}}$

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The remainder is not equal to zero hence (x+1) is not a factor of the given polynomial.