Given that (x-2) is a factor of the polynomial, we can use the Factor Theorem which states that a polynomial f(x) has a factor (x-a) if and only if f(a) = 0.

So, if (x-2) is a factor of P(x), then P(2) = 0.

Substitute x = 2 into the polynomial:

P(2) = (2)^4 - 3*(2)^3 + k*(2)^2 - 6*2 + 14 = 0
P(2) = 16 - 24 + 4k - 12 + 14 = 0
P(2) = -6 + 4k = 0

Solving for k, we get:

4k = 6
k = 6/4
k = 1.5

So, the value of k is 1.5.

Question

Given that (x-2) is a factor of the polynomial, we can use the Factor Theorem which states that a polynomial f(x) has a factor (x-a) if and only if f(a) = 0.

So, if (x-2) is a factor of P(x), then P(2) = 0.

Substitute x = 2 into the polynomial:

P(2) = (2)^4 - 3*(2)^3 + k*(2)^2 - 6*2 + 14 = 0
P(2) = 16 - 24 + 4k - 12 + 14 = 0
P(2) = -6 + 4k = 0

Solving for k, we get:

4k = 6
k = 6/4
k = 1.5

So, the value of k is 1.5.

Knowee AI · Accepted Answer

Given that (x-2) is a factor of the polynomial, we can use the Factor Theorem which states that a polynomial f(x) has a factor (x-a) if and only if f(a) = 0.

So, if (x-2) is a factor of P(x), then P(2) = 0.

Substitute x = 2 into the polynomial:

P(2) = (2)^4 - 3*(2)^3 + k*(2)^2 - 6*2 + 14 = 0
P(2) = 16 - 24 + 4k - 12 + 14 = 0
P(2) = -6 + 4k = 0

Solving for k, we get:

4k = 6
k = 6/4
k = 1.5

So, the value of k is 1.5.

P(x)=x4−3x3+kx2−6x+14𝑃(𝑥)=𝑥4−3𝑥3+𝑘𝑥2−6𝑥+14, where k is an unknown real number.If (x−2)(𝑥−2) is a factor of this polynomial, what is the value of k?

Question

P(x)=x^4−3x^3+kx^2−6x+14

Solution

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