# Problem of the Week

## Updated at Jan 9, 2023 11:10 AM

This week's problem comes from the equation category.

How would you solve the equation $$4(z-3)+{(4z)}^{2}=8$$?

Let's begin!

$4(z-3)+{(4z)}^{2}=8$

 1 Use Multiplication Distributive Property: $${(xy)}^{a}={x}^{a}{y}^{a}$$.$4(z-3)+{4}^{2}{z}^{2}=8$2 Simplify  $${4}^{2}$$  to  $$16$$.$4(z-3)+16{z}^{2}=8$3 Expand.$4z-12+16{z}^{2}=8$4 Move all terms to one side.$4z-12+16{z}^{2}-8=0$5 Simplify  $$4z-12+16{z}^{2}-8$$  to  $$4z-20+16{z}^{2}$$.$4z-20+16{z}^{2}=0$6 Factor out the common term $$4$$.$4(z-5+4{z}^{2})=0$7 Split the second term in $$z-5+4{z}^{2}$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$4\times -5=-20$2 Ask: Which two numbers add up to $$1$$ and multiply to $$-20$$?$$5$$ and $$-4$$3 Split $$z$$ as the sum of $$5z$$ and $$-4z$$.$4{z}^{2}+5z-4z-5$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$4(4{z}^{2}+5z-4z-5)=0$8 Factor out common terms in the first two terms, then in the last two terms.$4(z(4z+5)-(4z+5))=0$9 Factor out the common term $$4z+5$$.$4(4z+5)(z-1)=0$10 Solve for $$z$$.1 Ask: When will $$(4z+5)(z-1)$$ equal zero?When $$4z+5=0$$ or $$z-1=0$$2 Solve each of the 2 equations above.$z=-\frac{5}{4},1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$z=-\frac{5}{4},1$Done Decimal Form: -1.25, 1z=-5/4,1