Question

find the zeros of polynomial x^2-3x-m(m+3)



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Answer 1

Answer:

Step-by-step explanation:

f(x) = x2 – 3x – m (m + 3)  By adding and subtracting mx, we get  f(x) = x2 – mx – 3x + mx – m (m + 3)  = x[x – (m + 3)] + m[x – (m + 3)]  = [x – (m + 3)] (x + m)  f(x) = 0 ⇒ [x – (m + 3)] (x + m) = 0  ⇒ [x – (m + 3)] = 0 or (x + m) = 0  ⇒ x = m + 3 or x = –m  So, the zeroes of f(x) are –m and +3.

Answer 2

Step-by-step explanation:

Given : A polynomial $x^{2} -3x-m(m+3)$

To find : the zeroes of given polynomial.

Formula used :  If we have quadratic equation as,

⇒ $ax^{2} +bx+c=0$

Then

Discriminant of equation (D) is,

$D=b^{2} -4ac$

and zeroes are, $x1,x2=\frac{-b\pm\sqrt{D} }{2a}$

$x1=\frac{-b+\sqrt{D} }{2*a}$    and  $x2=\frac{-b-\sqrt{D} }{2*a}$.

a is the coefficient of $x^{2}$ , b is the coefficient of $x$ and c is for constant.

We know, to find the zeroes of the expression, we equate the expression with 0.

⇒ $x^{2} -3x-m(m+3)=0$

We can't use splitting method to split middle term because we have another equation as constant.

So we use discriminant formula for finding roots of equation.

Here in given equation we have,

$a=1$  ,  $b=-3$  and  $c=-m(m+3)$.

So $D=b^{2} -4ac$ is,

$D=(-3)^{2} -4*(1)*[-m(m+3)]$

   $=9+4(m)(m+3)$

   $=9+4m^{2} +12m$  

we get $D=4m^{2} +12m+9$ , a quadratic equation.

we can solve this equation by splitting middle term (finding factors of$(9*4=36)$ such that they add up to 12)

⇒ $4m^{2} +6m+6m+9=0$

⇒ $2m(2m+3)+3(2m+3)=0$

⇒ $(2m+3)^{2}$

we get discriminant (D) is $(2m+3)^{2}$.

now the zeroes are,

$x1,x2=\frac{-b\pm\sqrt{D} }{2a}$

           $=\frac{-(-3)\pm\sqrt{(2m+3)^{2} } }{2*1}$

          $=\frac{3\pm(2m+3)}{2}$

the x1 and x2 are,

$x1=\frac{-b+\sqrt{D} }{2*a}$                                                      $x2=\frac{-b-\sqrt{D} }{2*a}$

$x1=\frac{3+(2m+3)}{2}$                                                    $x2=\frac{3-(2m+3)}{2}$

    $=\frac{3+2m+3}{2}$                                                           $=\frac{3-2m-3}{2}$

    $=\frac{2m+6}{2}$                                                               $=\frac{-2m}{2}$

    $=\frac{2(m+3)}{2}$                                                            $=-m$  

     $=m+3$

The zeroes of given polynomial are $(m+3)$ and $-m$ .