Question

‌‌How‌ ‌do‌ ‌you‌ find ‌factor‌‌s ‌of ${{x}^{2}}-x-90$‌ ‌‌?‌

Answer 1

Hint: In quadratic equation $a{{x}^{2}}+bx+c$ we can factorize by finding 2 numbers such that their product is equal to $ac$ and their sum is equal to b. Then we can split the term $bx$ and factorize the quadratic equation.

Complete step-by-step solution:
The given equation is ${{x}^{2}}-x-90$ which is a quadratic equation. if we compare the equation to standard quadratic equation $a{{x}^{2}}+bx+c$ then $a=1$,$b=-1$ and $c=-90$
To factor a quadratic equation we can find two numbers m and n such that the sum of m and n is equal to b and the product of m and n is $ac$. Then we can split $bx$ to $mx+nx$ then we can factor the equation easily.
In our case
$\begin{align}
  & ac=1\times (-90) \\
 & ac=-90 \\
\end{align}$
and $b=-1$
So pair of 2 numbers whose product is -90 are (1,-90) , (2,-45) , (3,-30),(5,-18) , (9,-10), (10,-9),(15,-6),(30,-3) , (45,-2) and (90,-1) .
But there is only one pair whose sum is -1 that is (9,-10)
We can spilt $-x$ to $9x-10x$
So ${{x}^{2}}-x-90={{x}^{2}}+9x-10x-90$
Taking x common in the first half of the equation and taking -10 common in the second half of the equation.
$\Rightarrow {{x}^{2}}+9x-10x-90=x\left( x+9 \right)-10\left( x+9 \right)$
Taking $x+9$ common
$\Rightarrow x\left( x+9 \right)-10\left( x+9 \right)=\left( x+9 \right)\left( x-10 \right)$
In this way we can factor a quadratic equation; another method is to directly find the roots of the equation.

Note: While factoring a quadratic equation spilt the $bx$ term in such a way that the product of coefficient of x is equal to $ac$ (product of coefficient of ${{x}^{2}}$and constant term). Sometimes we can’t spilt $bx$ because the roots may come as irrational.in that case we have to fist find the roots of equation then we can write $a{{x}^{2}}+bx+c=a\left( x-\alpha \right)\left( x-\beta \right)$ where $\alpha $ and $\beta $ are roots of the equation.