Question

If one zero of the quadratic polynomial x^2-5x+k is -4 then value of k is ​

Answer 1

Answer :-

By using the factor theorem -

If -4 is a zero of the polynomial, then (x + 4) should be the factor of given polynomial.

So, dividing (x + 4) by the polynomial -

$\sf\Large\qquad\quad x - 9\\ \begin{array}{cc} \cline{2 - 2}\sf x + 4 )&\sf \ x^2 - 5x + k\\&\sf - x^2 - 4x \downarrow\\ \cline{2-2}& \sf \ \ \ \ -9x + k\\ &\sf \ +9x + 36 \\ \cline{2-2} & \sf \ \ k + 36 \\ \cline{2-2} \end{array}$

$\sf Divisor \rightarrow x + 4 \\ \\\sf Quotient \rightarrow x - 9\\\\\sf Remainder \rightarrow k + 36$

Here, we get the remainder as (k + 36)

As (x + 4) is a factor, remainder must be zero.

➻ $\sf k + 36 = 0$

➻ $\sf k = - 36$

We can also solve it another way -

As here we got the quotient as (x - 9) and the remainder must be zero, So we can say that ( x - 9 ) is a factor of the given polynomial.

So, now we get two factors - (x + 4) and (x - 9)

So,

➻ $\sf x^2 - 5x + k = ( x + 4 )( x - 9)$

➻ $\sf x^2 - 5x + k = x^2 + 4x - 9x - 36$

➻ $\sf \cancel{x^2} - \cancel{5x} + k = \cancel{x^2}-\cancel{5x} - 36$

➻ $\sf k = -36$

So, Value of k = -36

Answer 2

Answer:

Step-by-step explanation:

Given:

A quadratic polynomial : f(x)=x²-5x+k which is having -4 as a zero

To find :

The value of k

Solution

f(x)=x²-5x+k

As -4 is a zero of f(x)

so f(-4)=0

Putting x=-4 in f(x)

(-4)²-5(-4)+k=0

16+20+k=0

k = -36

Answer:

k = -36

Concept used:

If p is a zero of any polynomial f(x)

then f(p)=0

Other related concept:

1.If p is zero of f(x) then why f(p)=0?

Assuming f(x) = ax²+bx +c

By definition zero of a polynomial is the value of

variable x for which the value of the polynomial is zero

Here

p is a zero of f(x)

so f(p)=0

2. If α,β are zero of the polynomial

ax²+bx +c

Then

$\alpha + \beta =\frac{-b}{a}$

$\alpha \beta =\frac{c}{a}$

3. If Both zeros of f(x) are equal  then b² = 4ac