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Polynomials Quiz 2

mathematics

 The maximum number of zeroes, the polynomial $f\left( x \right) = {\left( {x - 3} \right)^2} + 9$ can have

 Maximum number of zeroes $ = $ Degree of the polynomial

 Maximum number of zeroes for a polynomial directly depends on the degree of the polynomial and in this case the degree is two, therefore two zeroes are possible.

 Find the value of $a\;{\text{and}}\;b$ if $2a - 3b = 4$ and also $\left( {x + 1} \right)$ is a factor of the polynomial $p\left( x \right) = 2{x^3} + a{x^2} + 2bx + 1$ 

 You have given factor of the given polynomial and eventually the zero of the polynomial, so put it in the polynomial and get another equation in $a\;{\text{and}}\;b$ then solve two equations of $a\;{\text{and}}\;b$ to get their values

Since $\left( {x + 1} \right)$ is a factor of the polynomial $p\left( x \right) = 2{x^3} + a{x^2} + 2bx + 1$ $ \\ $ Therefore $x =  - 1$ is a zero of the given polynomial that is $ \\  \Rightarrow p\left( { - 1} \right) = 0   \\   \Rightarrow 2{\left( { - 1} \right)^3} + a{\left( { - 1} \right)^2} + 2b\left( { - 1} \right) + 1 = 0   \\   \Rightarrow  - 2 + a - 2b + 1 = 0   \\   \Rightarrow a - 2b = 1 -  -  - (i)   \\  $ Also we have given $ \\ 2a - 3b = 4 -  -  - (ii) \\ $ So, performing equation $(ii) - 2 \times (i)$, we will get $ \\  \;\;2a - 3b = \;\;\;4   \\  \underline { - 2a + 4b =  - 2}    \\  \;\;\;\;\;\;\;\;\;\;\;b = \;\;\;2   \\  $  Putting $b = 2\;{\text{in}}\;(i)$ $ \\ \Rightarrow a - 2 \times 2 = 1   \\   \Rightarrow a = 1 + 4   \\ \Rightarrow a = 5   \\  $ Therefore $a = 5,\;b = 2$

 If a polynomial is given by $f\left( x \right) = a{x^2} + bx + c$ and $a + c = b$ then one of its zeroes is given as

 Use the given information to factorize the polynomial, substitute $b=a+c$ in the given quadratic equation.

We have $ \Rightarrow f\left( x \right) = a{x^2} + bx + c   \\   \Rightarrow f\left( x \right) = a{x^2} + \left( {a + c} \right)x + c\;\;\;\;\left[ {\because a + c = b} \right]   \\   \Rightarrow f\left( x \right) = a{x^2} + ax + cx + c   \\   \Rightarrow f\left( x \right) = ax\left( {x + 1} \right) + c\left( {x + 1} \right)   \\   \Rightarrow f\left( x \right) = \left( {x + 1} \right)\left( {ax + c} \right)   \\  $ Now, we have $\left( {ax + c} \right)$ as a factor of the given polynomial $ \\ $ Therefore $\left( {ax + c} \right) = 0 $  $ \\ \Rightarrow ax =  - c   \\   \Rightarrow x =  - \dfrac{c}{a}   \\  $ $x =  - \dfrac{c}{a}$ is one the zeroes of the given polynomial

 If $p\left( x \right) = a{x^2} + bx + c$ then the value of $\left( { - \dfrac{b}{a}} \right)\;{\text{and}}\;\dfrac{c}{a}$

 Divide the given polynomial with its leading coefficient so that the new leading coefficient becomes one and then compare it with the standard equation.

 Let us consider a quadratic polynomial $f\left( x \right) = {x^2} - \left( {d + e} \right)x + de$ and $p\left( x \right) = a{x^2} + bx + c \Leftrightarrow q\left( x \right) = {x^2} + \dfrac{b}{a}x + \dfrac{c}{a}$ $ \\  \Rightarrow f\left( x \right) = {x^2} - \left( {d + e} \right)x + de   \\   \Rightarrow f\left( x \right) = {x^2} - dx - ex + de   \\   \Rightarrow f\left( x \right) = x\left( {x - d} \right) - e\left( {x - d} \right)   \\   \Rightarrow f\left( x \right) = \left( {x - d} \right)\left( {x - e} \right)   \\  $ Here $d\;{\text{and}}\;e$ are the zeroes of $f\left( x \right)$ $ \\ $ So if we compare the two polynomials, $q\left( x \right)\;{\text{and}}\;f\left( x \right)$ $ \\ $ Then we will get that $ - \dfrac{b}{a} = \left( {d + e} \right)\;{\text{and}}\;\dfrac{c}{a} = de$ $ \\ $ That is $ - \dfrac{b}{a} = $ Sum of zeroes and $\dfrac{c}{a} = $ Product of zeroes

 Type of the zeroes of the quadratic polynomial $f\left( x \right) = {x^2} + 15x + 195$ are

 Find the zeros of the polynomial by factoring.

We have $f\left( x \right) = {x^2} + 15x + 195$ which cannot be segregated using the splitting the middle terms. $ \\ $ So, using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where, $b = 15$ and $c = 195$. $ \\ $ Substituting the values, we get $ \\  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}   \\   \Rightarrow x = \dfrac{{ - 15 \pm \sqrt {{{15}^2} - 4 \times 1 \times 195} }}{{2 \times 1}}   \\   \Rightarrow x = \dfrac{{ - 15 \pm \sqrt {225 - 780} }}{2}   \\   \Rightarrow x = \dfrac{{ - 15 \pm \sqrt { - 555} }}{2}   \\  $ Therefore its zeroes will be given as $ \\ x = \dfrac{{ - 15 + \sqrt { - 555} }}{2};x = \dfrac{{ - 15 - \sqrt { - 555} }}{2}$ which are imaginary.