Polynomials Test - 15

The number of zeroes of p(x) is the number of times the curve intersects the x-axis, i.e; attains the value $0$.

Here, the polynomial p(x) meets the x-axis at $4$ points. 

So, the number of zeroes $= 4$

Hence, the correct option is (C).

$x^3+2x^2-9x-18$ has $-2$ as root.

$x^3+2x^2-9x-18$

$=x^2\left(x+2\right)-9\left(x+2\right)$

$=\left(x^2-9\right)\left(x+2\right)$

$=\left(x+3\right)\left(x-3\right)\left(x+2\right)$   [using $a^2-b^2=\left(a+b\right)\left(a-b\right)$ ]

Hence, possible zeros of polynomial are $-2, +3$ and $-3$.

Thus, as $-2$ is already gives other possibles zeros of a polynomial are $+3$ and $-3$.

Hence, the correct option is (B).

Here, Divisor $=\left(x+1\right)$

Quotient $=3x-2$

Remainder $=3$

Dividend $=$ Divisor $\times$ Quotient $+$ Remainder

= $3x^2+3x-2x-2+3$

= $3x^2+x+1$

Hence, the correct option is (A).

From the graph, we get that the line is cutting the x-axis at 5 points.

∴ we can say that 5 zeroes will be represented by the polynomial.

Therefore, option A is correct.

From the equation, we get, the sum of roots = $\alpha +\beta +\gamma =-\frac{b}{a}$

and product of roots = $\alpha \times \beta \times \gamma =-\frac{d}{a}$

Hence, putting the values, we get $b=-5$ & $d=-3$.

Hence, Option C is the correct answer.

The sum of the roots of the polynomial is $=-\frac{\text{ coefficient of } x}{\text{ coefficient of } x^2}$

The product of the roots of the polynomial is $=\frac{\text{ constant term }}{\text{ coefficient of } x^2}$

Consider the given polynomial.

$25x^2+79x+12$

Sum of the zeroes of the polynomial= $-\frac{79}{25}$

Product of zeroes of the polynomial= $\frac{12}{25}$

∴ the product of the sum of zeroes and reciprocal of the product will be

$-\frac{b}{a}\times \frac{c}{a}$ = $-\frac{b}{a}\times \frac{a}{c}=-\frac{79}{25}\times \frac{25}{12}$

= $-\frac{79}{12}$

Hence, the correct option is (B).

We know that 

Dividend $=$ Divisor $\times$ Quotient $+$ Remainder...(1)

Here Dividend = $x^3+2x^2-4x+7$

Divisor $= x+2$

Quotient = $x^2-a$

Remainder $= 15$

$x^3+2x^2-4x+7=\left(x+2\right)\times \left(x^2-a\right)+15$

$\Rightarrow x^3+2x^2-4x+7=x^3-xa+2x^2-2a+15$

$\Rightarrow 7-4x=15-xa-2a$

$\Rightarrow xa+2a-4x=8$

$\Rightarrow xa+2a=8+4x$

$\Rightarrow a\left(x+2\right)=4\left(2+x\right)$

$\therefore a=4$

Hence, the correct option is (B).

$x^4-5x^2+4$ given roots are: $2$ and $-2$

$x^4-5x^2+4$

Let $x^2=t$

$\Rightarrow t^2-5t+4=0$

$\Rightarrow t^2-4t-t+4=0$

$\Rightarrow t\left(t-4\right)-1\left(t-4\right)=0$

$\Rightarrow \left(t-4\right)\left(t-1\right)=0$

$t=4, t=1$    [where $t=x^2$ ]

$x^2=4, x^2=1$

$x=\pm 2, x=\pm 1$

Hence, other two root are $\left(-1,1\right)$.

Hence, the correct option is (D).

As we know $\alpha +\beta +\gamma =-\frac{b}{a}$

According to question: $-\frac{b}{a}=-\left(\frac{-16}{4}\right)=4$

then, $\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{1}{4}$

Therefore, option A is correct.

As we know the general equation of the curve will be $a{x}^2-bx+c=0$

As the curve is cutting at $-3$ and $2$ on $x$-axis.

So, the equation that satisfies the graph will be

$x^2-\left(-3x+2\right)-6=0$

$x^2-\left(-x\right)-6=0$

$x^2+x-6=0$

Hence, the correct option is (A).

We know that 

Dividend = Divisor $\times$ Quotient + Remainder ..... (1)

Here Dividend =?

Divisor $=x + 2$

Quotient $=x^2 + 4x - 8$

Remainder $= 19$

Dividend $=(x+2)\times \left(x^2+4x-8\right)+19$

= $x^3+2x^2+4x^2+8x-8x-16+19$

= $x^3+6x^2+3$

Hence, the correct option is (C).

According to question $4x^3-4x^2-x+1$ = $0$

For the other possible zeroes, we will first take $-1$

So, we get $4x^3-4x^2-x+1 \ne 0$ (if we put $-1$ at the place of $x$).

This means that $-1$ could be one of the zeroes.

It also means that the number should be between $-1$ and $1$ from which we can get the zeroes.

Therefore, we will try $-\frac{1}{2}$ at the place of $x$.

$4x^3-4x^2-x+1=0$

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

$=4\times \left(-\frac{1}{8}\right)-4\times \left(\frac{1}{4}\right)+\left(\frac{1}{2}\right)+1$

$=\left(-\frac{4}{8}\right)-1+\frac{1}{2}+1$

$=-\frac{1}{2}-1+\frac{1}{2}+1$ (all the numbers will get eliminated and we will receive a zero).

$=0$

It means $-\frac{1}{2}$ is one of the zeroes.

like this for other zeroes, we will take one more number which is between -1 and 1 and check whether it is possible or not.

∴ $4x^3-4x^2-x+1=0$  (put $\frac{1}{2}$ as $x$)

Put $x=\frac{1}{2}$

$4x^3-4x^2-x+1=0$

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

$=4\times \frac{1}{8}-4\times \frac{1}{4}-\frac{1}{2}+1$

$=\frac{1}{2}-1-\frac{1}{2}+1$

$=0$

After solving similarly as done above we get the zero.

Therefore, we can say that $\frac{1}{2}$ and $-\frac{1}{2}$ will be the other possible zeroes.

Hence, the correct option is (D).

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

Sum of zeroes = $-\frac{b}{a}$

Product of zeroes $=\frac{c}{a}$

Sum of product of zeroes = $\left(\alpha +\beta \right) + \left(\alpha \times \beta \right) = -\frac{d}{a}$

Hence according to question:

Sum of roots: $\alpha +\beta =-\frac{b}{a}=-\frac{\text{ coefficient of }x}{\mathrm{coefficient} \text{of }{x}^2}$

Product of roots: $\alpha \beta =\frac{c}{a}=\frac{\text{ constant term }}{\text{ coefficient of }{x}^2}$

$b = -3, c = 24, d = 10$

Therefore, option D is correct.

As we can see from the graph, the general equation for this polynomial would be 

$y\left(x\right)=a{x}^2-bx+c$

the graph is cutting at -1 and -4 on the x-axis respectively.

Hence, obtaining two roots and at 4 on the y-axis.

Therefore, the obtained equation would be $x^2-\left(-x-4x\right)+4=0$

Resulting in the equation $x^2+5x+4=0$

Therefore, option A is correct.

From the graph given in the question, we get that the curve cuts the x-axis at two points i.e., $-2$ and $2$.

So, we will put the value of $-2$ in $y=a{x}^2+bx+c$

We get,

= $a{\left(-2\right)}^2+b\left(-2\right)+c$

= $4a^2-2b+c$ ....(1)

Now, putting $x=2$ in equation we get,

$a{x}^2+bx++c$

= $a{\left(2\right)}^2+b\left(2\right)+c = 4a^2+2b+c$ ....(2)

comparing (1) and (2) we get,

$-2b-2b=0$

i.e., $-4b=0$

Hence, $b = 0$

Therefore, option B is correct.