Question

Find the quadratic polynomial whose zeroes are \[ - 3\] and \[4\].

Answer 1

Hint: A quadratic equation in the variable \[x\] is an equation of the form of \[a{x^2} + bx + c = 0\]. We know that if \[p\left( x \right) = 0\], then we can say that \[x\] is the zero of the polynomials. In this question, we have given the zeros of the polynomial degree \[2\]. Firstly, we make both of them in factor form, then multiply to each other. After solving and re-arranging the terms, we get the quadratic equation in the form of \[a{x^2} + bx + c = 0\].

Complete step-by-step answer:
Step 1: Let the quadratic equation in variable \[x\]. So, we have given the zeros of equation \[ - 3\] and \[4\] respectively,
that is, \[x = - 3\] and \[x = 4\]
Re-arranging both the terms
i.e., \[x + 3 = 0\] and \[x - 4 = 0\]
Step 2: We can write the equation, \[\left( {x + 3} \right)\left( {x - 4} \right) = 0\]. It means, in which both terms, one of them is zero, that is, \[x + 3 = 0\] or \[x - 4 = 0\]
After multiplying these terms, we get
\[ \Rightarrow x\left( {x - 4} \right) + 3\left( {x - 4} \right) = 0\]
\[ \Rightarrow {x^2} - 4x + 3x - 12 = 0\]
Re-arranging the terms and after solving the terms \[ - 4x\] and \[3x\], we get
\[\left( {x + 3} \right)\left( {x - 4} \right) = {x^2} - x - 12 = 0\]
\[ \Rightarrow {x^2} - x - 12 = 0\] is the form of a quadratic equation in variable \[x\].
Hence, \[{x^2} - x - 12\] is quadratic polynomial whose zeroes are \[ - 3\] and \[4\].

Note: In general, if \[\alpha \] and \[\beta \] are the zeroes of the quadratic polynomial \[p\left( x \right) = a{x^2} + bx + c,a \ne 0\], then we know that \[\left( {x - \alpha } \right)\] and \[\left( {x - \beta } \right)\] are the factors of \[p\left( x \right)\].
Therefore, \[a{x^2} + bx + c = k\left( {x - \alpha } \right)\left( {x - \beta } \right)\], where \[k\] is a constant
\[ = k\left[ {{x^2} - \left( {\alpha + \beta } \right)x + \alpha \beta } \right]\]
\[ = k{x^2} - k\left( {\alpha + \beta } \right)x + k\alpha \beta \]
Comparing the coefficient of \[{x^2},x\] and constant terms on both sides, we get \[a = k,b = - k\left( {\alpha + \beta } \right)\] and \[c = k\alpha \beta \]
This gives \[\alpha + \beta = \dfrac{{ - b}}{a}\] and \[\alpha \beta = \dfrac{c}{a}\].