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Sagot :
Given:
The trinomial is
[tex]c^2+2c+n[/tex]
To find:
The value of n for which the given trinomial is a perfect square and then factor the trinomial.
Solution:
We know that, the perfect square trinomial is given by
[tex](a+b)^2=a^2+2ab+b^2[/tex] ...(i)
We have,
[tex]c^2+2c+n[/tex] ...(ii)
From (i) and (ii), we get
[tex]a^2=c^2, 2ab=2c, b^2=n[/tex]
[tex]a=c[/tex]
Similarly,
[tex]2ab=2c[/tex]
[tex]2ab=2a[/tex] [tex][\because a=c][/tex]
[tex]b=\dfrac{2a}{2a}[/tex]
[tex]b=1[/tex]
Now,
[tex]n=b^2[/tex]
[tex]n=1^2[/tex]
[tex]n=1[/tex]
The value of n is 1.
The trinomial is
[tex]c^2+2c+1[/tex]
Splitting the middle term, we get
[tex]=c^2+c+c+1[/tex]
[tex]=c(c+1)+1(c+1)[/tex]
[tex]=(c+1)(c+1)[/tex]
[tex]=(c+1)^2[/tex]
Therefore, the factorized form of the trinomial is [tex](c+1)^2[/tex].
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