Question

Find k, so that x² + 2x + k is a factor of 2x⁴ + x³ – 14x² + 5x + 6. Also, find all the zeroes of the two polynomials.

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Answer 1

Answer:

Since x² + 2x + k is a factor of 2x⁴ + x³ – 14x² + 5x + 6, so we dividing 2x⁴ + x³ – 14x² + 5x + 6 by x²+2x+k should leave the remainder 0 as shown in the attachment:

Comparing the coefficient of x, we get

$\sf \: 21+7k=0$

$\sf \implies \: 7k=−21$

$\sf \implies \: k= \cancel\frac{ - 21}{7}$

$\sf \implies \: k = - 3$

The equation x²+2x+k becomes x²+2x−3 and the factors of x²+2x−3=0 are: x²+3x−x−3=0

$\sf \: x(x+3)−1(x+3)=0$

$\sf \implies \: (x−1)(x+3)=0$

$\sf \implies \: x=1,−3$

Now, the equation 2x²−3x−(8+2k) becomes 2x²−3x−2 and the factors of 2x² −3x−2=0 are:

$\sf \: 2x^2 −4x+x−2=0$

$\sf \implies \: 2x(x−2)+1(x−2)=0$

$\sf \implies \: (2x+1)(x−2)=0$

$\sf \implies \: x= \frac{ - 1}{2} ,2$

Hence, the zeros of the given two polynomials are 1,−3,− 1/2 and 2.

Answer 2

Step-by-step explanation:

Here in this question, concept of factorisation as well as long division is used. We are given 2 polynomials where one is a factor of other. We are asked to find the zeroes of both the polynomials and also the value of k. Since $x^2 + 2x + k$ is a factor of $2x^4 + x^3 - 14x^2 + 5x + 6$, the polynomial when divided by $x^2 + 2x + k$ should leave a remainder as 0.

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Let's perform long division.

[See attachment]

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Since the remainder is $(21 + 7k)x + 2k^2 +8k + 6$, but it should be equal to 0 (as we have discussed above).

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${:\implies (21 + 7k)x +2k^2 + 8k+6=0}$

Coefficient of $x$ should be equal to 0.

${:\implies 21 + 7k = 0}$

${:\implies 7k = -21}$

Divide by 7 from both sides

${:\implies 7k\div 7 = -21 \div 7}$

${:\implies k = -3}$

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So the required value of k is -3.

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$\rule{200}{1}$ ‎

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Now we have to find factors of polynomials.

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Factors of x² + 2x +k are :-

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${:\implies x^2 + 2x +k = 0}$

Substitute value of k = -3

${:\implies x^2 + 2x -3 = 0}$

${:\implies x^2 +3x - x -3 = 0}$

${:\implies x(x + 3) -1(x +3) = 0}$

${:\implies (x+3)(x-1)= 0}$

${:\implies x = -3\: and\: 1}$

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$\rule{200}{1}$ ‎ ‎

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Factors of 2x² - 3x -(8+2k) [Dividend] are :-

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${: \implies 2x^2 -3x - (8+2k)=0}$

Put k = -3

${: \implies 2x^2 -3x - (8+2(-3))=0}$

${: \implies 2x^2 -3x - (8-6)=0}$

${: \implies 2x^2 -3x - 2=0}$

${: \implies 2x^2 -4x +x- 2=0}$

${: \implies 2x(x-2)+1(x-2)=0}$

${: \implies (2x+1) (x-2)}$

${: \implies x = 2, \dfrac{-1}{2}}$

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$\rule{200}{1}$ ‎

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Factors of 2x⁴ + x³ - 14x² + 5x + 6 :-

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Since polynomial (x²+2x+k) times (2x²-3x-(8+2k)) = 2x⁴ + x³ - 14x²+5x+6, the factors of 2x⁴ + x³ - 14x²+5x+6 will be the factors of [(x²+2x+k) times (2x²-3x-(8+2k))].

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So the factors of 2x⁴ + x³ - 14x² +5x + 6 are:-

• 2, -1/2, -3 and 1.

$\rule{200}{1}$ ‎

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