Question

How do you factor completely $4{{x}^{2}}-12x+8$ ?

Answer 1

Hint: When we factorize a quadratic equation $a{{x}^{2}}+bx+c$ we try to find 2 integers m and n such that sum of m and n is equal to b and product of m and n is product of a and c. Then we can bx as mx+nx then factorize the equation.

Complete step by step answer:
The given equation is $4{{x}^{2}}-12x+8$ which is a quadratic equation. if we compare the equation to standard quadratic equation $a{{x}^{2}}+bx+c$ then a=4, b= -12 and c= 8
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
$ac=32$
And b= -12
So pair of 2 numbers whose product is 32 are (1,32), (2,16) , (4,8) , (-8,-4)
But there is only one pair whose sum is -12 that is (-8,-4)
We can -12x split to -8x-4x
So Taking x common in the first half of the equation and taking 1 common in the second half of the equation.
$\Rightarrow 4{{x}^{2}}-12x+8=4{{x}^{2}}-8x-4x+8$
Taking 4x common in first half of the equation and -4 common in second half of the equation
$\Rightarrow 4{{x}^{2}}-12x+8=4x\left( x-2 \right)-4\left( x-2 \right)$
Taking x -2 common form equation we get
$\Rightarrow 4{{x}^{2}}-12x+8=\left( 4x-4 \right)\left( x-2 \right)$
Now we can take 4 common from 4x-4
$\Rightarrow 4{{x}^{2}}-12x+8=4\left( x-1 \right)\left( x-2 \right)$
 In this way we can factor a quadratic equation; another method is directly find the roots of
the equation.

Note:
Sometimes we can’t factorize by using above method due to the roots of the quadratic equation may be fraction or irrational number; in that case we can find the roots by quadratic formula and directly write $a{{x}^{2}}+bx+c=a\left( x-\alpha \right)\left( x-\beta \right)$ where $\alpha ,\beta $ are the roots of the equation.