
How do you factor quadratic equations with a coefficient?
Answer
547.8k+ views
Hint: The general form of a quadratic equation is \[a{{x}^{2}}+bx+c\]. To express a quadratic equation in its factored form. We have to find its roots. Say \[\alpha ,\beta \] are the two real roots of the equation. Then the factored form is \[a\left( x-\alpha \right)\left( x-\beta \right)\]. We can find the roots of the equation using the formula method as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. As we already know that, if \[x=a\] is a root of a polynomial function, then \[x-a\] is one of its factors and vice versa.
Complete step by step solution:
Let’s take an example of a quadratic equation having a coefficient.
We take the quadratic expression \[5{{x}^{2}}+34x+24\], we need to factorize it. On comparing with the general solution of the quadratic equation \[a{{x}^{2}}+bx+c\], we get \[a=5,b=34\And c=24\].
To express in factored form, we first have to find the roots of the equation \[5{{x}^{2}}+34x+24\].
We can find the roots of the equation using the formula method.
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of the coefficients in the above formula, we get
\[\begin{align}
& \Rightarrow x=\dfrac{-(34)\pm \sqrt{{{\left( 34 \right)}^{2}}-4(5)(24)}}{2(5)} \\
& \Rightarrow x=\dfrac{-34\pm \sqrt{676}}{10} \\
& \Rightarrow x=\dfrac{-34\pm 26}{10} \\
\end{align}\]
\[\Rightarrow x=\alpha =\dfrac{-34+26}{10}=\dfrac{-8}{10}\] or \[x=\beta =\dfrac{-34-26}{10}=\dfrac{-60}{10}\]
Canceling out the common factor from both numerator and denominator for the above fraction, we get
\[\Rightarrow \alpha =\dfrac{-4}{5}\] and \[\beta =-6\]
Now that, we have the roots of the given expression, we can express it as its factored form as follows,
For the quadratic expression \[5{{x}^{2}}+34x+24\], \[a=5\] and the roots as \[\alpha =\dfrac{-4}{5}\And \beta =-6\].
The factored form is,
\[\begin{align}
& \Rightarrow 5\left( x-\left( -\dfrac{4}{5} \right) \right)\left( x-\left( -6 \right) \right) \\
& \Rightarrow 5\left( x+\dfrac{4}{5} \right)\left( x+6 \right) \\
\end{align}\]
Note: A polynomial expression can have the maximum number of factors equal to the degree of the polynomial. For quadratic equations, the degree is two. Hence, the maximum number of factors it can have is two.
Complete step by step solution:
Let’s take an example of a quadratic equation having a coefficient.
We take the quadratic expression \[5{{x}^{2}}+34x+24\], we need to factorize it. On comparing with the general solution of the quadratic equation \[a{{x}^{2}}+bx+c\], we get \[a=5,b=34\And c=24\].
To express in factored form, we first have to find the roots of the equation \[5{{x}^{2}}+34x+24\].
We can find the roots of the equation using the formula method.
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of the coefficients in the above formula, we get
\[\begin{align}
& \Rightarrow x=\dfrac{-(34)\pm \sqrt{{{\left( 34 \right)}^{2}}-4(5)(24)}}{2(5)} \\
& \Rightarrow x=\dfrac{-34\pm \sqrt{676}}{10} \\
& \Rightarrow x=\dfrac{-34\pm 26}{10} \\
\end{align}\]
\[\Rightarrow x=\alpha =\dfrac{-34+26}{10}=\dfrac{-8}{10}\] or \[x=\beta =\dfrac{-34-26}{10}=\dfrac{-60}{10}\]
Canceling out the common factor from both numerator and denominator for the above fraction, we get
\[\Rightarrow \alpha =\dfrac{-4}{5}\] and \[\beta =-6\]
Now that, we have the roots of the given expression, we can express it as its factored form as follows,
For the quadratic expression \[5{{x}^{2}}+34x+24\], \[a=5\] and the roots as \[\alpha =\dfrac{-4}{5}\And \beta =-6\].
The factored form is,
\[\begin{align}
& \Rightarrow 5\left( x-\left( -\dfrac{4}{5} \right) \right)\left( x-\left( -6 \right) \right) \\
& \Rightarrow 5\left( x+\dfrac{4}{5} \right)\left( x+6 \right) \\
\end{align}\]
Note: A polynomial expression can have the maximum number of factors equal to the degree of the polynomial. For quadratic equations, the degree is two. Hence, the maximum number of factors it can have is two.
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