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Sagot :
Answer:
[tex]\frac{1}{4}[/tex] (2x + 11)²
Step-by-step explanation:
given
x²+ 11x + [tex]\frac{121}{4}[/tex]
Convert the coefficients of the x²- term and the x- term to fractions with denominator of 4, that is
1 = [tex]\frac{4}{4}[/tex] and 11 = [tex]\frac{44}{4}[/tex]
Rewrite the expression as
[tex]\frac{4}{4}[/tex] x² + [tex]\frac{44}{4}[/tex] x + [tex]\frac{121}{4}[/tex] ← factor out [tex]\frac{1}{4}[/tex] from each term
= [tex]\frac{1}{4}[/tex] (4x² + 44x + 121 )
factorise 4x² + 44x + 121
Rewrite 4x² as (2x)² and 121 as 11²
= (2x)² + 44x + 11²
Consider the perfect square trinomial
a² + 2ab + b² = (a + b)²
Now check that the middle term 44x is two times the product of the numbers being squared in the first and third terms
2 × 2x × 11 = 44x
Then with a = 2x and b = 11
4x² + 44x + 121 = (2x + 11)²
Finally
x² + 11x + [tex]\frac{121}{4}[/tex] = [tex]\frac{1}{4}[/tex] (2x + 11)²
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