Polynomials Test - 20

If the graph of a polynomial intersects the x-axis at three points, then the number of zeroes are 3 because the x- axis is intersect three times by the three coordinates so,

Number of zeroes of the polynomial =  number of the coordinates of the points (where its graph intersects the x-axis)

Therefore, the correct option is (C).

Given polynomial p(x) = ax + b

Put p(x) = 0

$\Rightarrow$ ax + b = 0

$\Rightarrow$ax =−b

$\Rightarrow$x = −ba

Hence, the correct option is (A).

In each points (-2,0), (0,0) and (2,0) all in x- axis and each points intersect the graph of cubic polynomials.

So, the number zeroes are -2,0 and 2 (which is coordinates of x-axis)

Hence, the correct option is (A).

The zero of the polynomial is defined as any real value of  for which the value $x$ of the polynomial becomes zero.

A real number $k$ is a zero of a polynomial $p\left(x\right)$, if $p\left(k\right)=0$

For example $P\left(x\right)=x^2-3x-4$

Then $P(-1)=(-1{)}^2-(3\times -1)-4=0$

and $P\left(4\right)=(4{)}^2-(3\times 4)-4=0$

For a polynomial $P\left(x\right)$, real number $k$ is said to be zero of polynomial $P\left(x\right)$, if $P\left(k\right) = 0$.

Hence, the correct option is (D).

If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial $a{x}^2+bx+c$,

$\because$ Sum of the zeroes of a quadratic polynomial

$a{x}^2+bx+c=\frac{-\left(\text{ coefficient of }x\right)}{\text{ coefficient of }{x}^2}$ then $\alpha +\beta =\frac{-b}{a}$

Hence, the correct option is (C).

$f\left(x\right)=2x-5$ line which intersects the $x$ axis at exactly one point.

At $x$ axis $y=0$

$f\left(x\right)=2x-5$

$y=2x-5$

$0=2x-5$

$x=\frac{5}{2}$

Hence points are $(\frac{5}{2},0)$.

Therefore, the correct option is (B).

It is given that the roots of a quadratic polynomial are \(\frac 14\) and \(-1\).

Let \(\alpha=\frac 14\) and \(\beta = -1\)

Sum of roots \(=(\alpha+\beta)\)

\(=\frac 14-1\)

\(=-\frac34\)

Product of roots \(=\alpha\beta\)

\(=\frac 14\times-1\)

\(=-\frac 14\)

We know that, equation of quadratic polynomial of roots \(\alpha\) and \(\beta\) is given by,

\(x^{2}-(\alpha+\beta) x+(\alpha \beta)=0\)

Therefore,

\(x^{2}-\left(-\frac{3}{4}\right) x+\left(-\frac{1}{4}\right)=0\)

\(\Rightarrow x^{2}+\frac{3}{4} x-\frac{1}{4}=0\)

\(\Rightarrow \frac{4 x^{2}+3 x-1}{4}=0\)

\(\Rightarrow 4 x^{2}+3 x-1=0\)

Hence, the correct option is (A).

(x² + 9x + 20) = 0  Splitting the middle term, we get

x² + 5x + 4x + 20 = 0

= x(x+5) + 4(x+5) = 0

= (x+5) (x+4) = 0

∴x+5 = 0 and x + 4 = 0

⇒ x = −5 and x = −4

Hence, the correct option is (B).

If $\alpha ,\beta$ and $\gamma$ are the zeroes of a cubic polynomial $a{x}^3+b{x}^2+cx+d$,

$\because$ Sum of the zeroes of a cubic polynomial $a{x}^3+b{x}^2+cx+d=\frac{-\left(\text{ coefficient of }{x}^2\right)}{\text{ coefficient of }{x}^3}$, then $\alpha +\beta +\gamma =\frac{-b}{a}$

Hence, the correct option is (B).

$\alpha , \beta$ and $\gamma$ are the zeroes of a cubic polynomial $a{x}^3+b{x}^2+cx+d$,

Therefore, Sum of the product of zeroes of a cubic polynomial $a{x}^3+b{x}^2+cx+d=\frac{\left(\text{ coefficient of }x\right)}{\text{ coefficient of }{x}^3}$, then $\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$

Hence, the correct option is (D).

If $\alpha , \beta$ and $\gamma$ are the zeroes of a cubic polynomial $a{x}^3+b{x}^2+cx+d$,

$\because$ Sum of the product of zeroes of a cubic polynomial $a{x}^3+b{x}^2+cx+d=\frac{-\text{ constant term }}{\text{ coefficient of }{x}^3}$, then $\alpha \beta \gamma =\frac{-d}{a}$

Hence, the correct option is (A).

Let one zero of the given polynomial be $\alpha$

The other zero is reciprocal be $\frac{1}{\alpha }$.

$\because$ Product of roots $\alpha$ and $\frac{1}{\alpha }=\frac{c}{a}$

$\alpha \times \frac{1}{\alpha }=\frac{3k}{k+4}$

$1=\frac{3k}{k+4}$

$k+4=3k$ (by cross multiplication)

$4=3k-k$

$4=2k$

$k=\frac{4}{2}$

$k=2$

Hence, the correct option is (D).

If $\alpha$ and $\beta$ are the zeroes of a quadratic polynomial $a{x}^2+bx+c$,

$\because$ Product of the zeroes of a quadratic polynomial $a{x}^2+bx+c=\frac{\text{ constant term }}{\text{ coefficient of }{x}^2}$, then $\alpha \beta =\frac{c}{a}$

Hence, the correct option is (A).

 x²- (Sum the Zeroes ) x + (Product of Zeroes) 

x² - (-4)x + 3

=  x² + 4x + 3 

Hence, the correct option is (C).

Let one zeroes of the given polynomial be $\alpha$ and $\beta$.

According to the question, $\alpha +\beta =0$

$\frac{-b}{a}=0$

$\frac{-(p+3)}{2}=0$

$\Rightarrow -(p+3)=0$ (by cross multiplication)

$\Rightarrow p=-3$

Hence, the correct option is (B).