I always shot to use the factorizing an approach first. However sometimes the an approach does not work-related for specific quadratic equations. In that case, I will to utilizing the quadratic formula$$x=\frac-b \pm \sqrtb^2-4ac2a$$For her equation that $2x^2+7x-15=0$, we have $a=2,b=7,c=-15$.

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If you can factorize your quadratic without making use of the formula then you need to do it, since it is normally faster.

When you have actually a quadratic $ax^2+bx+c$ you can easily factorize that if you can discover two number $n_1,n_2$ such that $n_1+n_2=b$ and $n_1n_2=ac$ through rewriting $bx$ as $n_1x+n_2x$ and also then grouping similar terms.

In the case of $2x^2+7x-15=0$ we have actually $n_1=10$ and also $n_2=-3$ so us factorize the without having to usage the quadratic formula:

$2x^2+7x-15=0$$2x^2+10x-3x-15=0$$2x(x+5)-3(x+5)=0$$(2x-3)(x+5)=0$

**Key Idea**$\ $ use AC-method to mitigate to factoring a polynomial that"s $\,\rm\color#c00monic\,$ (lead coeff $=\color#c001)$

$$\quad\ \ \begineqnarrayf &\,=\,& \ \ 2\ x^2+\ 7\ x\ -\,\ 15\\\Rightarrow\ 2f &\,=\,&\ (2x)^2\! +7(2x)-30\\ &\,=\,& \ \ \ \color#c00X^2+\,7\ X\ -\,\ 30,\,\ \ X\, =\, 2x\\ &\,=\,& \ \ \,(X-3)\ (X+\,10)\\ &\,=\,& \ \ (2x-3)\,(2x+10)\\\Rightarrow\ f\,=\, 2^-1(2f) &\,=\,& \ \ (2x - 3)\,(x+5)\\\endeqnarray\qquad\qquad$$

It"s my view that the quadratic formula should always be offered to factor a quadratic. That"s what it"s for.

Completing the square must be used to change a quadratic into vertex-focus form (and to derive the quadratic formula).

Factoring by group is only really teach at the HS level to provide students an development to Number theory (finding integer remedies to equations, evaluating factors and residuals, etc), or possibly as a preliminary for emerging strategies for factoring cubics. Other than that, it"s not very useful.

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