How do you factor the trinomial $10g^2+10=29g$ ?

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Answer 1 · 2016-01-30T17:25:38

This is an equation, and so I'll solve the equation for you.

$10g^2 + 10 = 29g$

$10g^2 - 29g + 10 = 0$

When factoring trinomials of the form $ax^2 + bx + c$ , where $a!=0,1$ , you must find two trinomials that multiply to $a xx c$ and that add to $b$

Two numbers that multiply to $100$ and add to $- 29$ are $- 25$ and $- 4$

$10g^2 - 4g - 25g + 10 = 0$

$2g(5g - 2) - 5(5g - 2) = 0$

$(2g - 5)(5g - 2) = 0$

Set each of the individual parts equal to $0$ and solve both of them.

$g = 5/2$ and $2/5$

So, your solution set is $g = 5/2$ and $2/5$

How do you factor the trinomial 10g^2+10=29g10g^2+10=29g?