rocedwrp

2021-03-08

Find the roots of the function
$f\left(x\right)=\left(2x-1\right)\cdot \left({x}^{2}+2x-3\right)$, with $x\in R$

### Answer & Explanation

StrycharzT

First find the factors of $\left({x}^{2}+2x-3\right).$
Compare the $a{x}^{2}+bx+c$ and ${x}^{2}+2x-3$, implies that $a=1,b=2$, and $c=-3$.
The ac-method of factoring quadratics: To find factors of quadratic $a{x}^{2}+bx+c$, Find two numbers whose sum is b and product is $a\cdot c$.
Here $a\cdot c=1\cdot \left(-3\right)=-3$and $b=2$.
To find factors of quadratic ${x}^{2}+2x-3$, find two numbers whose sum is 2 and the product is -3.
Such numbers are 3 and -1.
The factors of $\left({x}^{2}+2x-3\right)$ are $\left(x+3\right)\left(x-1\right).$
The given function $f\left(x\right)=\left(2x-1\right)\cdot \left({x}^{2}+2x-3\right)$ becomes
$f\left(x\right)=\left(2x-1\right)·\left(x+3\right)·\left(x-1\right)$
Set $f\left(x\right)=0$, implies that
$\left(2x-1\right)\cdot \left(x+3\right)\cdot \left(x-1\right)=0$
By using zero product property,
$\left(2x-1\right)=0,\left(x+3\right)=0,\left(x-1\right)=0$
Implies that,
$2x=1,x=-3,x=1$
That is, the roots of the given function are,
$x=\frac{1}{2},x=-3,x=1.$

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