How do you factor the expression $n^{2} - 28 n + 196$ $n^2 - 28n + 196$ ?

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Answer 1 · 2016-03-05T16:52:56

$(n - 14) (n - 14)$ $color(blue)( (n-14) (n-14)$ is the factorised form of the expression.

$n^{2} - 28 n + 196$ $n^2-28n+196$

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like $a n^{2} + b n + c$ $an^2 + bn + c$ , we need to think of 2 numbers such that:

$N_{1} \cdot N_{2} = a \cdot c = 1 \cdot 196 = 196$ $N_1*N_2 = a*c = 1*196 = 196$

AND

$N_{1} + N_{2} = b = - 28$ $N_1 +N_2 = b = -28$

After trying out a few numbers we get $N_{1} = - 14$ $N_1 = -14$ and $N_{2} = - 14$ $N_2 =-14$

$- 14 \cdot - 14 = 196$ $-14*-14 = 196$ , and $(- 14) + (- 14) = - 28$ $(-14)+(-14)= -28$

$n^{2} - 28 n + 196 = n^{2} - 14 n - 14 n + 196$ $n^2-28n+196 = n^2-14n - 14n+196$

$= n (n - 14) - 14 (n - 14)$ $= n (n-14) -14(n-14)$

$(n - 14) (n - 14)$ $color(blue)( (n-14) (n-14)$ is the factorised form of the expression.

How do you factor the expression n^2 - 28n + 196n2−28n+196n^2 - 28n + 196?