Polynomials Test - 19

Polynomial $p\left(x\right)=x^2-9=0$

$\Rightarrow x^2-3^2=0$

According the formula

$a^2-b^2=(a+b)(a-b)$

$\Rightarrow (x+3)(-3)=0$

So, 

$\Rightarrow (x+3)=0$ or $(x-3)=0$

$x=-3$ or $x=3$

Zeroes are  $(3,-3)$

Sum of zeroes are: $3-3=0$

Hence, the correct option is (C).

$x^2-\left(\sqrt{2}+\sqrt{3}\right)x+\sqrt{6}$

Sum of zeroes = $\frac{-b}{a}=-\frac{\text{ coefficient of }x}{\text{ coefficient of }{x}^2}$

= $\frac{-\left(\sqrt{2}+\sqrt{3}\right)}{1}$

= $\sqrt{2}+\sqrt{3}$

Therefore, option D is correct.

Sum of zeroes = $\frac{-b}{a}$ = $\frac{-\left(-1\right)}{1}$ = $\frac{1}{1}=1$

product of zeroes = $\frac{c}{a}=\frac{-2}{1} = -2$

So, the equation will be = $a{x}^2+bx+c$

$\left(1\right)x^2+\left(-1\right)x+\left(-2\right)$

$= x^2-x-2$

Therefore, option C is correct.

The curve is cutting the x-axis three times having 3 zeroes resulting in the degree of equation 4.

Therefore, option B is correct.

We have $x^4+x^3+2x^2+3x$

$\Rightarrow x^4+2x^2+x^3+2x+x$ (Break $3x$ as $2x+x$ )

$\begin{array}{l}\Rightarrow x^2\left(x^2+2\right)+x\left(x^2+2\right)+x=\left(x^2+x\right)\left(x^2+2\right)+x\end{array}$

On dividing by $x^2+x$, we get

$(x^2+2)+\frac{x}{x^2+x}$ (It is of form $Q+\frac{R}{D}$)

So remainder is $x$.

Hence, the remainder is $x$.

Therefore, option B is correct.

Since the curve doesn't touch the x-axis at any point.

The number of zeroes will be 0.

Therefore, option A is correct.

Since the curve is touching at x-axis at a time.

So, the zeroes will be 1.

Therefore, option C is correct.

As the curve cuts the x-axis 10 times

it has 10 zeroes.

Therefore, option B is correct.

as we know that Dividend = divisor $\times$ quotient + remainder

then, $x^4+2x^2-3x+17 = g\left(x\right) \times \left(x^2+5\right) + \left(-3x+22\right)$

 $x^4+2x^2-3x+7 - (-3x+22) = g\left(x\right) \times x^2+5$

$x^4+2x^2-3x+7+3x-22= g\left(x\right) \times x^2+5$

$\Rightarrow \frac{x^4+2x^2-15}{x^2+5}=g\left(x\right)$

$\Rightarrow x^2-3 = g\left(x\right)$

$a{x}^2+bx+b=0$

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

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

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

$\Rightarrow \frac{{\displaystyle \frac{-b}{a}}}{\sqrt{{\displaystyle \frac{b}{a}}}}+\sqrt{\frac{b}{a}}$

$\Rightarrow -\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{a}}$

= 0

Therefore, option A is correct.

There are three curves given u, v, and w.

As we can see in the graph the curve u cuts the line at 3 points.

Whereas, v and w curves clearly cuts the line at 2-2 points respectively. 

So, from the above information, we clearly get that u have the greatest number of real and distinct zeroes.

Therefore, option A is correct.

Sum of zeroes = $\frac{-b}{a}$$=\frac{-11}{2}=-\frac{\text{ coefficient of }x}{\text{ coefficient of }{x}^2}$

Product of zeroes = $\frac{c}{a} = \frac{5}{2}$ $=\frac{\text{ constant term }}{\text{ coefficient of }{x}^2}$

The general equation will be = $a{x}^2+bx+c = 0$

The equation will be = $2x^2-11x+5=0$

Therefore, option A is correct.

Let the three roots of $x^3+7x^2-2x-14$ be $\alpha ,\beta ,\gamma$ so,

$\alpha =\sqrt{2},\beta =-\sqrt{2}$ and $\gamma =?$

We know that sum of roots $(\alpha +\beta +\gamma )$ for a cubic polynomial is $a{x}^3+b{x}^2+cx+d$ is $\frac{-b}{a}$

So here $\alpha +\beta +\gamma =\frac{-b}{a}=\frac{(-7)}{1}=-7$

$\alpha +\beta +\gamma =-7$...(i)

Also $\alpha =\sqrt{2}, \beta =\sqrt{2}$, so substituting in equation (i)

$\left(\sqrt{2}\right)+(-\sqrt{2})+\gamma =-7$

$\Rightarrow \gamma =-7$

Hence, the third zero of the given polynomial is $-7$.

Therefore, option B is correct.

$x^3+2x^2-3x-1=0$

$\alpha +\beta +\gamma =-2$

$\alpha \beta +\beta \gamma +\gamma \alpha = -3$

and $\alpha \beta \gamma =1$

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2} = \frac{{\beta }^2{\gamma }^2+{\alpha }^2{\gamma }^2+{\alpha }^2{\beta }^2}{{\alpha }^2{\beta }^2{\gamma }^2}$

$\Rightarrow \frac{{\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)}^2-2\alpha \beta \gamma \left(\alpha +\beta +\gamma \right)}{{\left(\alpha \beta \gamma \right)}^2}$

$\Rightarrow \frac{{\left(-3\right)}^2-2\times 1\left(-2\right)}{\left(1\right)} = 9+4 = 13$

Therefore, option C is correct.

The curve is cutting the x-axis four times having 4 zeroes resulting in the degree of the equation 5.

Therefore, option C is correct.