How to factor the polynomial $2x^2-7x-15$?

Question

Having difficulties in factoring the expressions $2x^2 - 7x -15$.

Answer 1

We can split the linear term if we can find two numbers with product $2 \cdot -15 = -30$ and sum $-7$. The factorizations of $-30$ are \begin{align*} -30 & = 1 \cdot -30 & -30 & = -1 \cdot 30\\ & = 2 \cdot -15 & & = -2 \cdot 15\\ & = \color{blue}{3 \cdot -10} & & = -3 \cdot 10\\ & = 5 \cdot -6 & & = -5 \cdot 6 \end{align*} Of the eight pairs of factors only, $3$ and $-10$ have sum $-30$. Hence, \begin{align*} 2x^2 - 7x - 15 & = 2x^2 + 3x - 10x - 15 & \text{split the linear term}\\ & = x(2x + 3) - 5(2x + 3) & \text{factor by grouping}\\ & = (x - 5)(2x + 3) & \text{extract the common factor} \end{align*} Note that when you compute the product $(x - 5)(2x + 3)$, you perform the steps of the factorization by splitting and grouping in reverse order.

Why does this work?

Suppose $ax^2 + bx + c$ factors with respect to the rational numbers as $$ax^2 + bx + c = (rx + s)(tx + u)$$ Then \begin{align*} ax^2 + bx + c & = (rx + s)(tx + u)\\ & = rx(tx + u) + s(tx + u)\\ & = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\ & = \color{blue}{rt}x^2 + (\color{green}{ru} + \color{green}{st})x + \color{blue}{su} \end{align*} Observe that the product of the coefficients of the quadratic and constant terms is equal to the product of the coefficients of the two terms that sum to the coefficient of the linear term, that is, $$(\color{blue}{rt})(\color{blue}{su}) = (\color{\green}{ru})(\color{green}{st}) = rstu$$ Matching coefficients yields $a = rt$, $b = ru + st$, and $c = su$. Hence, we can find a factorization with respect to the rationals if there exist two numbers with product $ac$ and sum $b$. Those two numbers are the numbers we use to split the linear term.

Answer 2

The first terms will be $2x$ and $x$. The last terms will probably be $3$ and $5$. So you have one of these: $$(2x\pm3)(x\pm5)\\(2x\pm5)(x\pm3)$$ Now just go through all of them until the $x$ term is $-7x$.

Answer 3

For $2x^2-7x-15=0$

$x=\dfrac{7\pm\sqrt{7^2-4\cdot2(-15)}}{2\cdot2}=5,-\dfrac32$

$$\implies2x^2-7x-15=2\left(x-\left(-\dfrac32\right)\right)(x-5)=(2x+3)(x-5)$$

Generalization :

For $a\ne0,$ $$ax^2+bx+c=\dfrac{(2ax+b)^2-(\sqrt{b^2-4ca})^2}{4a}=\cdots$$