How to factor and solve quadratic equations step by step

Key concept: These problems all use the same idea: move everything to one side, factor completely, then use the zero product property. Once that pattern clicks, the individual equations become much easier to handle.

Step by step

24. $x^2+2x+1=0$

Factor:

$x^2+2x+1=(x+1)^2$

So

$x+1=0 \Rightarrow x=-1.$

25. $x^2-5x-14=0$

Factor:

$x^2-5x-14=(x-7)(x+2).$

So

$x=7 \quad \text{or} \quad x=-2.$

26. $x^2+7x=0$

Factor out $x$:

$x(x+7)=0.$

So

$x=0 \quad \text{or} \quad x=-7.$

27. $2x^2-5x+2=0$

Factor:

$2x^2-5x+2=(2x-1)(x-2).$

So

$x=\frac12 \quad \text{or} \quad x=2.$

28. $2x^2+3x=5$

Move all terms to one side:

$2x^2+3x-5=0.$

Factor:

$2x^2+3x-5=(2x+5)(x-1).$

So

$x=-\frac52 \quad \text{or} \quad x=1.$

29. $5x^2+16x=-3$

Move $-3$ to the left:

$5x^2+16x+3=0.$

Factor:

$5x^2+16x+3=(5x+1)(x+3).$

So

$x=-\frac15 \quad \text{or} \quad x=-3.$

Shaded expression

The prompt says “Write an equation to represent the shaded,” but the shaded diagram is not included here. If you send the figure, I can translate it into an equation directly.

Pitfall alert

Two common slips: forgetting to move every term to one side before factoring, and factoring correctly but then missing one solution. Also watch signs carefully; a small sign error changes the whole pair of roots.

Try different conditions

If a quadratic does not factor nicely, you can still solve it using the quadratic formula. For example, if you changed #27 to $2x^2-5x+1=0$, factoring is harder, so you would use

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

Further reading

quadratic equation, factoring, zero product property