Question

Question 20

Factorised form of $r^2-10r+21$ is
a) $(r-1)(r-4)$
b) $(r-7)(r-3)$
c) $(r-7)(r+3)$
d) $(r+7)(r+3)$

Answer 1

b) $(r-7)(r-3)$

We have,
​​​​​​​ $r^2-10r+21=r^2-7r-3r+21=r(r-7)-3(r-7)$
[by spliting the middle term, so that product of their numerical coefficients is equal constant term]
$=(r-7)(r-3)$ $[\because x^2+(a+b)x+ab=(x+a\left)\right(x+b\left)\right]$