Polynomials
POLYNOMIALS
Factorisation of Quadratic polynomial by splitting the middle term
Solving a Quadratic equation \(
ax^2 + bx + c = 0\)
Step I : Splitting of middle term
i) First multiply a and c i.e., ac (or) -ac
ii) Split ‘b’ into factors as their product is ac (or) -ac
Step II : Let the factors of \(
ax^2 + bx + c = (px + q)(rx + s)\)
\(
\Rightarrow px + q = 0\) (or) \(
rx + s = 0\)
\(
\Rightarrow x = \frac{{ - q}}
{p}\) (or) \(x=\frac {-s}
{r}\)
Ex : factorize \(
2x^2 + 9x + 10\)
Here \(
ac = 2 \times 10 = 20
\)
\(
b = 9 = 4 + 5\)
\(
ac = 4 \times 5 = 20
\)
Factorisation of the difference of two squares
1) \(
(a + b)^2 - (a - b)^2 = 4ab\)
2) \(
ab = \left( {\frac{{a + b}}
{2}} \right)^2 - \left( {\frac{{a - b}}
{2}} \right)^2 \)
Ex : \(
(x + y + 2z)(x + y)\) as the difference of 2 squares
\(
(x + y + 2z)(x + y) = \left( {\frac{{(x + y + 2z) + (x + y)}}
{2}} \right)^2 - \left( {\frac{{(x + y + 2z) - (x + y)}}
{2}} \right)^2 \)
\(
= \left( {\frac{{2x + 2y + 2z}}
{2}} \right)^2 - \left( {\frac{{2z}}
{2}} \right)^2 \)
\(
(x + y + 2z)(x + y) = (x + y + z)^2 - z^2 \)