Select all polynomials that have (x – 3) as a factor.

There's a much faster way that doesn't involve complicated division. Recall that if (x-k) is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.

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Here's a fairly short proof:

Consider a polynomial q(x) such that

p(x) = (x-k)q(x)

which shows that (x-k) is a factor of p(x). We don't need to worry about what q(x) actually is since it will go away effectively.

If we plug x = k into the p(x) function, we get

p(x) = (x-k)q(x)

p(k) = (k-k)q(k)

p(k) = 0*q(k)

p(k) = 0

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How is this useful? Well we can note that the factor (x-3) is in the form (x-k) where k = 3.

So the idea is to plug x = 3 into each of the four functions and see which result in 0. If we get 0 as an output, then we have a factor. Otherwise, it's not a factor.

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Plug x = 3 into the first function

A(x) = x^3 - 2x^2 - 4x + 3

A(3) = 3^3 - 2(3)^2 - 4(3) + 3

A(3) = 0

This shows (x-3) is a factor of the function A(x)

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Repeat for the second function

B(x) = x^3 + 3x^2 - 2x - 6

B(3) = 3^3 + 3(3)^2 - 2(3) - 6

B(3) = 42

We don't get an output of 0, so (x-3) cannot be a factor of B(x)

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Now onto the third function

C(x) = x^4 - 2x^3 - 27

C(3) = 3^4 - 2(3)^3 - 27

C(3) = 0

So (x-3) is a factor of C(x)

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Finally the last function

D(x) = x^4 - 20x - 21

D(3) = 3^4 - 20(3) - 21

D(3) = 0

Therefore (x-3) is a factor of D(x) as well.

oeerivona oeerivona · Answer 2 · 2021-10-19T13:36:14+02:00

The remainder theorem will be used to identify the polynomial that has a remainder of zero when divided by (x - 3)

The polynomials that have 0 as a factor are;

(C) C(x) = x⁴ - 2·x³ - 27

(D) D(x) = x⁴ - 20·x - 21

Reason:

According to the remainder theorem, we have;

The remainder following the division of a polynomial, by (x - a) is P(a)

Therefore, (x - 3) is a factor of the polynomial P(x) where P(3) = 0

For option (A), we have; A(x) = x³ - 2·x² - 4·x + 3

The remainder when (x³ - 2·x² - 4·x + 3) is divided by (x - 3) is therefore;

A(3) = 3³ - 2 × 2² - 4 × 2 + 3 = 14

Therefore, (x - 3) is not a factor of A(x) = x³ - 2·x² - 4·x + 3

For option (B), we have;

B(x) = x³ + 3·x² - 2·x - 6

The remainder when B(x) is divided by (x - 3) is therefore;

B(3) = 3³ + 3×3² - 2×3 - 6 42

Therefore, (x - 3) is not a factor of B(x) = x³ + 3·x² - 2·x - 6

For option (C), we have;

C(x) = x⁴ - 2·x³ - 27

The remainder when C(x) is divided by (x - 3) is therefore;

C(3) = 3⁴ - 2×3³ - 27 = 0

Therefore;

(x - 3) is a factor of C(x) = x⁴ - 2·x³ - 27

For option (D), we have;

D(x) = x⁴ - 20·x - 21

The remainder when D(x) is divided by (x - 3) is given as follows;

D(3) = 3⁴ - 20×3 - 21 = 0

Therefore;

(x - 3) is a factor of D(x) = x⁴ - 20·x - 21

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