What is the quadratic polynomial function in standard form with zeros \[3\], \[1\], \[2\] and \[ - 3\] ?
Answer
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Hint: We have to find the quartic polynomial function in standard form with zeros \[3\], \[1\], \[2\] and \[ - 3\]. We will consider zeroes as the roots for getting the required factors by assuming the polynomial is equal to zero. Then we will multiply the factors to get the required polynomial. At last, we will rearrange the polynomial in the decreasing power of \[x\], to get the polynomial in standard form.
Complete step by step answer:
The zeroes of the polynomial are given as \[3\], \[1\], \[2\] and \[ - 3\]. This means they are the roots when the polynomial is equated to zero. So, if \[3\], \[1\], \[2\] and \[ - 3\] are the four roots of a polynomial than the factors are \[\left( {x - 3} \right)\], \[\left( {x - 1} \right)\], \[\left( {x - 2} \right)\] and \[\left( {x + 3} \right)\]. By multiplying all the factors, we get the polynomial as \[\left( {x - 3} \right)\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)\].
Taking two at a time and on multiplying, we get
\[\left( {{x^2} - x - 3x + 3} \right)\left( {{x^2} + 3x - 2x - 6} \right)\]
\[\Rightarrow \left( {{x^2} - 4x + 3} \right)\left( {{x^2} + x - 6} \right)\]
Now, on multiplying the terms, we get
\[ {x^4} + {x^3} - 6{x^2} - 4{x^3} - 4{x^2} + 24x + 3{x^2} + 3x - 18\]
\[\therefore {x^4} - 7{x^2} - 3{x^3} + 27x - 18\]
For the standard form, we should write the polynomial in the decreasing power of \[x\]. So, we get the polynomial as \[{x^4} - 3{x^3} - 7{x^2} + 27x - 18\].
Therefore, the quartic polynomial function in standard form with zeros \[3\], \[1\], \[2\] and \[ - 3\] is \[{x^4} - 3{x^3} - 7{x^2} + 27x - 18\].
Note: A quadratic equation, is an equation of degree four that equates a quartic polynomial to zero, of the form \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\], where \[a\] is not equal to zero. The derivative of quartic function is a cubic function. Sometimes, the term biquadratic is used instead of quartic, but usually, biquadratic function refers to a quartic polynomial without terms of odd degree.
Complete step by step answer:
The zeroes of the polynomial are given as \[3\], \[1\], \[2\] and \[ - 3\]. This means they are the roots when the polynomial is equated to zero. So, if \[3\], \[1\], \[2\] and \[ - 3\] are the four roots of a polynomial than the factors are \[\left( {x - 3} \right)\], \[\left( {x - 1} \right)\], \[\left( {x - 2} \right)\] and \[\left( {x + 3} \right)\]. By multiplying all the factors, we get the polynomial as \[\left( {x - 3} \right)\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)\].
Taking two at a time and on multiplying, we get
\[\left( {{x^2} - x - 3x + 3} \right)\left( {{x^2} + 3x - 2x - 6} \right)\]
\[\Rightarrow \left( {{x^2} - 4x + 3} \right)\left( {{x^2} + x - 6} \right)\]
Now, on multiplying the terms, we get
\[ {x^4} + {x^3} - 6{x^2} - 4{x^3} - 4{x^2} + 24x + 3{x^2} + 3x - 18\]
\[\therefore {x^4} - 7{x^2} - 3{x^3} + 27x - 18\]
For the standard form, we should write the polynomial in the decreasing power of \[x\]. So, we get the polynomial as \[{x^4} - 3{x^3} - 7{x^2} + 27x - 18\].
Therefore, the quartic polynomial function in standard form with zeros \[3\], \[1\], \[2\] and \[ - 3\] is \[{x^4} - 3{x^3} - 7{x^2} + 27x - 18\].
Note: A quadratic equation, is an equation of degree four that equates a quartic polynomial to zero, of the form \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\], where \[a\] is not equal to zero. The derivative of quartic function is a cubic function. Sometimes, the term biquadratic is used instead of quartic, but usually, biquadratic function refers to a quartic polynomial without terms of odd degree.
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