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# Find values of p,q and r: 3x^2+12x+5=p(x+q)^2 Watch

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1. I'm stuck, any tips or advice would be appreciated.
2. 3x^2+12x+5
(3x)^2+12x+5
9x+12x+5
21x+5
//////////////////
Where is R?
3. (Original post by SNK0)
3x^2+12x+5
(3x)^2+12x+5
9x+12x+5
21x+5
//////////////////
Where is R?
AHHHHHHHHHHH I FORGOT THE R.
3x^2+12x+5=p(x+q)^2+r
4. (Original post by AxSirlotl)
AHHHHHHHHHHH I FORGOT THE R.
3x^2+12x+5=p(x+q)^2+r
= p (x^2 + 2xq + q^2) + r
= px^2 + 2xpq + pq^2 + r

Comparing this with 3x^2 + 12x + 5, we may write:

p = 3

2pq = 12 --------------> q = 2

pq^2 + r = 5---------> r = -7
5. (Original post by SNK0)
= p (x^2 + 2xq + q^2) + r
= px^2 + 2xpq + pq^2 + r

Comparing this with 3x^2 + 12x + 5, we may write:

p = 3

2pq = 12 --------------> q = 2

pq^2 + r = 5---------> r = -7
Okay I've got it, thanks a lot!
6. It's basically expansion.

convert p(x+q)^2+r to the form of 3x^2+12x+5

p(x+q)^2 = p(x+q)(x+q)

p(x+q)(x+q) = p(x^2 + xq + q^2 + xq)

p(x^2 + xq + q^2 + xq) = p(x^2 + 2xq + q^2)

p(x^2 + 2xq + q^2) = px^2 + 2pxq + pq^2

px^2 + 2pxq + pq^2 + r

3x^2 + 12x + 5

px^2 = 3x^2
p = 3

2pxq = 12
2(3x+q) = 6x+q = 12x
q = 2

pq^2 + r = 3(2)^2 + r = 12 + r
r = -7

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Updated: October 9, 2016
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