Factorizing Quadratic Expressions by Completing the Square

The expression is $x^2+10x+3$, so $b=10$ and $c=3$.

$\frac{10}{2}=5$, so the squared bracket is $(x+5{)}^2$.

We must subtract $5^2=25$.

$(x+5{)}^2-25+3=(x+5{)}^2-22$.

$(x+5{)}^2-22=x^2+10x+25-22=x^2+10x+3$ ✓

Half of $10$ is $5$, so the bracket must be $(x+5{)}^2$. The option $(x+10{)}^2-97$ uses the full coefficient, not half — eliminate it.

$(x+5{)}^2+3$ has not subtracted $5^2=25$ at all — it just kept $+3$. Eliminate it.

$(x+5{)}^2-28$ gives $-28$, but $-25+3=-22$, not $-28$. Eliminate it.

$(x+5{)}^2-22$ expands to $x^2+10x+25-22=x^2+10x+3$ ✓

The $x$-coefficient is $b=7$.

$\frac{7}{2}$ is half of $7$.

${\left(\frac{7}{2}\right)}^2=\frac{49}{4}$.

The constant that must be subtracted is $\frac{49}{4}$.

$49=7^2$ — this is the square of the full coefficient, not half of it. Incorrect.

$\frac{7}{2}$ is half the coefficient but it has not been squared yet. We need the square of this value.

$\frac{7}{4}$ is simply $\frac{7}{2}÷2$, not ${\left(\frac{7}{2}\right)}^2$. Incorrect.

${\left(\frac{7}{2}\right)}^2=\frac{49}{4}$ is the correct constant to subtract.