The Null Factor Law
Null Factor Law
Fill in the blanks.
$6\times \square =0$
$\left(-3\right)\times \square =0$
$a\times \square =0$
$\square \times 7=0$
$\square \times \left(-5\right)=0$
$\square \times b=0$
Fill in the blank. If $a\times b=0$, then $a=\square$ or $b=\square$ or both.
Fill in the blanks.
$6\times \left(\square-5\right) =0$
$\left(-3\right) \left(\square +4\right) =0$
$a\times \left(\square -p\right) =0$
$\left(\square +9\right) \times 7=0$
$\left(\square -\frac{1}{2}\right) \left(-5\right)=0$
$\left(\square -q\right) \times b=0$
Solve for $x$. If there are multiple answers, separate them with commas.
$\left(x-2\right)\left(x-1\right)=0\qquad \qquad x=$
$\left(x-7\right)\left(x+7\right)=0\qquad \qquad x=$
$\left(x-4\right)^2=0\qquad \qquad x=$
$x\left(x+8\right)=0\qquad \qquad x=$
$3\left(x-5\right)\left(x+4\right)=0\qquad \qquad x=$
$-2\left(x+6\right)^2=0\qquad \qquad x=$
Solve for $x$. If there are multiple answers, separate them with commas.
$\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\qquad \qquad x=$
$\left(x-7\right)\left(x+7\right)^2=0\qquad \qquad x=$
$\left(x-5\right)^3=0\qquad \qquad x=$
$x^2\left(x+8\right)\left(x-\frac{1}{2}\right)=0\qquad \qquad x=$
$-\left(x+2\right)\left(x-3\right)^2\left(x+1\right)^3=0\qquad \qquad x=$