The sum of zeroes of the polynomial f(x) = 2³- 3² + 4x – 5 is 6, then the value of k is

a) 2

b) -2

The sum of zeroes of the polynomial f(x) = 2³- 3² + 4x – 5 is 6, then the value of k is

a) 2

b) -2

c) 4

d) -4
Answer is ASAP​

2 thoughts on “The sum of zeroes of the polynomial f(x) = 2³- 3² + 4x – 5 is 6, then the value of k is<br /><br />a) 2<br /><br />b) -2<br /><br”

  1. QUESTION:

    • The sum of zeroes of the polynomial f(x) = 2x³- 3kx² + 4x – 5 is 6, then the value of k is?

    ANSWER:

    Given:

    • f(x) = 2x³- 3kx² + 4x – 5
    • Sum of zeroes = 6

    To Find:

    • Value of k

    Solution:

    [tex]\text{We are given that,}\\\:\longrightarrow f(x)=2x^3-3kx^2+4x-5\\\\\text{We know that,}\\\\\text{For a cubic polynomial, $p(x)=ax^2+bx^2+cx+d$,}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{\text{Coefficient of x$^2$}}{\text{Coefficient of x$^3$}}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{b}{a}\\\\\text{So,}\\\\\text{Here, a = 2 and b = -3k. So,}\\\\:\implies\text{Sum of zeroes}=-\dfrac{-3k}{2}\\\\:\implies\text{Sum of zeroes}=\dfrac{3k}{2}[/tex]

    [tex]\text{But, we are given that,}\\\\:\longrightarrow\text{Sum of zeroes}=6\\\\\text{So,}\\\\:\implies\dfrac{3k}{2}=6\\\\\text{On cross-multiplying,}\\\\:\implies3k=6\times2\\\\:\implies3k=12\\\\\text{On transposing 3 to RHS,}\\\\:\implies k=\dfrac{12\!\!\!\!\!/^{\:\:4}}{3\!\!\!/}\\\\\bf{:\implies k=4}\\\\\text{\bf{Hence, value of ‘k’ is 4(Option C).}}[/tex]

    Formula Used:

    [tex]\text{For a cubic polynomial, $p(x)=ax^2+bx^2+cx+d$,}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{\text{Coefficient of x$^2$}}{\text{Coefficient of x$^3$}}\\\\:\hookrightarrow\text{Sum of zeroes}=-\dfrac{b}{a}[/tex]

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