Negative Sign Before Trinomial

[Jason76]

Considering: \(\displaystyle ax^{2} + bx + c\)

When factoring \(\displaystyle -14x^{2} + 39x - 10\), how to get rid of the negative sign? I know how to do the rest, you simply find two numbers whose product \(\displaystyle = a * c\) and sum \(\displaystyle = b\)

 

[pka]

Considering: \(\displaystyle ax^{2} + bx + c\)
When factoring \(\displaystyle -14x^{2} + 39x - 10\), how to get rid of the negative sign?

Factor \(\displaystyle -(14x^2-39x+10)\)

 

[Deleted member 4993]

Considering: \(\displaystyle ax^{2} + bx + c\)

When factoring \(\displaystyle -14x^{2} + 39x - 10\), how to get rid of the negative sign? I know how to do the rest, you simply find two numbers whose product \(\displaystyle = a * c\) and sum \(\displaystyle = b\)

In many ways:

[FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]14[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]39[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]10[/FONT] = = [FONT=MathJax_Main]−([/FONT][FONT=MathJax_Main]14[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]-[/FONT][FONT=MathJax_Main]39[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]10)[/FONT]

Another way to handle that would be to assign a = - 14

so in this case the product to look for is = (-14)*(-10) = 140

And sum to look for is = 39

In this case those numbers are 4 & 35

so

-14x² + 39x -10 = -14x² + 35x + 4x - 10 = -7x(2x - 5) + 2(2x - 5) = (2x - 5)(-7x +2) = -(2x - 5)(7x - 2)

similarly:


-14x² + 39x -10 = -(14x² - 39x +10) = -(14x² - 35x - 4x + 10) = -[7x(2x - 5) - 2(2x - 5)] = -(2x - 5)(7x - 2)

 

[Jason76]

Sounds good. Thanks.