Question 1209216
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If -5 and 5 are the roots of this monic polynomial, then (x+5) and (x-5) are factors


(x+5)(x-5) = x^2 - 25 due to the difference of squares rule.


Therefore the <font color=red>answer</font> is:
<pre>x^2 + <u> <font color=red>0</font> </u>x + <u> <font color=red>-25</font> </u></pre>
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Another approach.


Let's say the blanks are b and c for now.


We have x^2+bx+c


If p,q are the roots of this quadratic, then (x-p) and (x-q) are factors.


(x-p)(x-q) = x^2-qx-px+pq
(x-p)(x-q) = x^2-(p+q)x+pq


We have 
x^2+bx+c = x^2-(p+q)x+pq


Comparing terms shows that 
-(p+q)x = bx which leads to b = -(p+q)
and,
c = pq


In short,
b = -(p+q)
c = pq
Which are part of Vieta's Formulas.


In this case the roots are p = -5 and q = 5
b = -(p+q) = -(-5+5) = <font color=red>0</font>
c = p*q = -5*5 = <font color=red>-25</font>


We arrive at the <font color=red>answer</font>  
<pre>x^2 + <u> <font color=red>0</font> </u>x + <u> <font color=red>-25</font> </u></pre>
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