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This is one of our many quizzes on Polynomial roots. Click on this to access more such competitive quizzes on Algebra.

Polynomial roots Quiz 2

mathematics

 If $Rs.2x$ is given to Ram and double the amount of Ram and $Rs.70$ less is given to Shivam. Write the total amount given to both of them as a polynomial expression.

 To calculate the total sum, add the amount given to Ram and Shivam's amount. Note that Shivam has twice the amount of Ram.

Amount given to Ram $Rs.2x$ $ \\ $ Give, Amount is given to Shivam $ = Rs.\,({(2x)^2} - 70) = Rs.(4{x^2} - 70)$ $ \\ $ So, the total amount given to both $ = Rs.2x + (4{x^2} - 70) = Rs.(4{x^2} + 2x - 70)$

 If the sum of the roots of a quadratic expression is 8 and the product of the roots is 15. Then write the quadratic expression belonging to the roots.

 The general form of a quadratic expression is $a{x^2} + bx + c$ $ \\ $ Sum of the roots $ =  - \dfrac{b}{a}$ $ \\ $ Product of the roots $ = \dfrac{c}{a}$ .

W.K.T, $a{x^2} + bx + c$. $ \\ $ Given, Sum of the roots $ = 8 =  - \dfrac{{( - 8)}}{1} =  - \dfrac{b}{a}$ $ \\ $ Product of the roots $ = 15 = \dfrac{{15}}{1} = \dfrac{c}{a}$ $ \\ $ From the above given info. $ \\ $ $a = 1;b =  - 8\,and\,c = 15$ $ \\ $ Now according to the general form $ \\ $ The required quadratic expression is ${x^2} - 8x + 15$ .

 Check whether $ - \dfrac{4}{3}$ is the root of the polynomial $3{x^2} + 10x + 8$ ?

 Substitute $ - \dfrac{4}{3}$ in the given polynomial. If the value is not equal to 0, then it is not the root of a given polynomial.

To check $ - \dfrac{4}{3}$ is the root of the given polynomial or not, we should substitute -4 in the place of $x$ and  if the concluded value of polynomial is equals to 0 , then -4 is the root of given polynomial. $ \\ $ But, after substituting, the concluded value will not equal to zero, that means -4 is not the root of the given polynomial. $ \\  \therefore 3{x^2} + 10x + 8 \\   = 3{( - \dfrac{4}{3})^2} + 10( - \dfrac{4}{3}) + 8 \\   = 3(\dfrac{{16}}{9}) - \dfrac{{40}}{3} + 8 \\   = \dfrac{{16}}{3} - \dfrac{{40}}{3} + 8 \\   = \dfrac{{16 - 40 + 24}}{3} \\   = \dfrac{0}{3} = 0 \\  $ Therefore, after substitution, the value of the polynomial is 0. So, $ - \dfrac{4}{3}$ is  the root of a given polynomial $3{x^2} + 10x + 8$ .

 If the length of a rectangle is $4x$ and the perimeter is $9x$ . Then find the area of the rectangle.

 Assume the breadth of the rectangle first. Then find the breadth with the help of given perimeter and length. And then find the area.

Let the breadth of the rectangle be $b$ $ \\ $ Given, Length of the rectangle is $4x$ $ \\ $ Perimeter of the rectangle is $9x$ Now,  $ \\ \text{Perimeter of rectangle }= 2\text{(length + breadth)} \\   \Rightarrow 9x = 2(4x + b) \\   \Rightarrow 9x = 8x + 2b \\   \Rightarrow 2b = 9x - 8x \\   \Rightarrow b = \dfrac{x}{2} \\  $ So, the breadth of the rectangle is $\dfrac{x}{2}$ . $ \\ $ Area of the rectangle: $\text{Area of the rectangle }=\text{ length }\times\text{ breadth} \\ = 4x \times \dfrac{x}{2} = 2{x^2} \\  $

 If the area of the rectangle is $3{x^2} - 14x + 8$ . Then find the perimeter of the rectangle.

 Factorise the given area. Factors are length and breadth. 

Given, Area of the rectangle is $3{x^2} - 14x + 8$ $ \\ $ Factorise the area: $ \\ 3{x^2} - 14x + 8 \\   \Rightarrow 3{x^2} - 12x - 2x + 8 \\   \Rightarrow 3x(x - 4) - 2(x - 4) \\   \Rightarrow (x - 4)(3x - 2) \\  $ So, the length and breadth of the rectangle are $(x - 4)\,and\,(3x - 2)$ $ \\ $ Perimeter: $ \\ \text{Perimeter of the rectangle }= 2\text{(length + breadth)} \\ = 2(x - 4 + 3x - 2) \\ = 2(4x - 6) = 8x - 12 \\  $