You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction to Factorization

Question 20Fa...

Question

Question 20

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

Open in App

Solution

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

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

Suggest Corrections

Similar questions

r^{2} - 10 r + 21

(r - 1) (r - 4)

(r - 7) (r - 3)

(r - 7) (r + 3)

(r + 7) (r + 3)

\infty \sum r = 1 \frac{r}{7^{r}}

n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \dots \infty]

7^{r}

55!

r =

A B C, r = \frac{R}{6}

r_{1} = 7 r

A =

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program