Factorised form of $r^{2} - 10 r + 21$ is:

$(r - 1) (r - 4)$
$(r - 7) (r - 3)$
$(r - 7) (r + 3)$
$(r + 7) (r + 3)$

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Solution

The correct option is B
$(r - 7) (r - 3)$

Factorise the given equation:
The given expression is $r^{2} - 10 r + 21$
Factorise by splitting the middle term as follows:
$\begin{array}{rcl} r^{2} - 10 r + 21 & = & r^{2} - 7 r - 3 r + 21 \\ = & r (r - 7) - 3 (r - 7) \\ = & (r - 3) (r - 7) \\ = & (r - 7) (r - 3) \end{array}$
Hence option (B) is the correct answer.

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Similar questions

r^{2} - 10 r + 21

(r - 1) (r - 4)

(r - 7) (r - 3)

(r - 7) (r + 3)

(r + 7) (r + 3)

r^{2} - 10 r + 21

(r - 1) (r - 4)

(r - 7) (r - 3)

(r - 7) (r + 3)

(r + 7) (r + 3)

p (q^{2} + r^{2}) - q (r^{2} + p^{2})

Which of the following is the factorised form of $20 l^{2} m + 30 a l m ?$

(p^{2} - 2 p q + q^{2}) - r^{2}

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