factoring help
- Thread starter ptebwwong
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Why guess? Test it: \(\displaystyle 80^{2} - 4 \cdot 25 \cdot 81 = -1700\) -- No Real solutions. Certainly no Rational Solutions.
It cannot be factored with Rational numbers or with Integers.
Also, x = -(-80)/(2*25) = 8/5. This is the low point of the graph of y = 25x^2 - 80x + 81. What is the value of y at x = 8/5?
It cannot be factored with Rational numbers or with Integers.
Also, x = -(-80)/(2*25) = 8/5. This is the low point of the graph of y = 25x^2 - 80x + 81. What is the value of y at x = 8/5?
One little trick you can use to see if its factorable is check the discriminant.
It it is a perfect square then it is factorable.
\(\displaystyle b^{2}-4ac\)
\(\displaystyle (-80)^{2}-4(25)(81)=-1700\)
Not a perfect square and not factorable. Also, it is negative. That means it has non-real solutions.
Suppose we had \(\displaystyle 25x^{2}+15x-54\)
The discriminant is \(\displaystyle 15^{2}-4(25)(-54)=b^2-4ac=5625=75^{2}\). A perfect square, so it's factorable.
\(\displaystyle 25x^{2}+15x-54=(5x-6)(5x+9)\)
It it is a perfect square then it is factorable.
\(\displaystyle b^{2}-4ac\)
\(\displaystyle (-80)^{2}-4(25)(81)=-1700\)
Not a perfect square and not factorable. Also, it is negative. That means it has non-real solutions.
Suppose we had \(\displaystyle 25x^{2}+15x-54\)
The discriminant is \(\displaystyle 15^{2}-4(25)(-54)=b^2-4ac=5625=75^{2}\). A perfect square, so it's factorable.
\(\displaystyle 25x^{2}+15x-54=(5x-6)(5x+9)\)