think about x^2-2x-3. Variable the expression by grouping. First, the expression needs to it is in rewritten as x^2+ax+bx-3. To discover a and also b, collection up a mechanism to it is in solved.

You are watching: Factor completely −3x2 + 6x − 9.


Since abdominal is negative, a and b have actually the the opposite signs. Because a+b is negative, the negative number has higher absolute value than the positive. The just such pair is the system solution.
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displaystyleleft(x-3 ight)^2-18 Explanation: displaystylex^2-6x-9=left(x^2-6x+9 ight)-18displaystylex^2-6x+9=left(x-3 ight)left(x-3 ight)=left(x-3 ight)^2 ...
x2-6x-9=1 Two services were discovered : x =(6-√76)/2=3-√ 19 = -1.359 x =(6+√76)/2=3+√ 19 = 7.359 Rearrange: Rearrange the equation by subtracting what is come the best of the equal sign from ...
3x2-26x-9 Final result : (x - 9) • (3x + 1) step by action solution : action 1 :Equation in ~ the end of step 1 : (3x2 - 26x) - 9 step 2 :Trying to element by separating the middle term ...
displaystyle=3left(x+1 ight)left(x-3 ight) Explanation: displaystyle3x^2-6x-9displaystyle=3left(x^2-2x-3 ight)(taking the end the ...
ns think you expected to compose 3x^2−6x−9=0 . Climate your first steps are fine: 3x^2−6x=9 3x(x-2)=9 in ~ this point, ns think girlfriend erroneously be separate this equation right into two equations 3x=9quadmathrmor ...
3x2-6x-45 Final an outcome : 3 • (x + 3) • (x - 5) action by step solution : step 1 :Equation at the finish of step 1 : (3x2 - 6x) - 45 step 2 : step 3 :Pulling out favor terms : 3.1 Pull out ...
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Consider x^2-2x-3. Factor the expression by grouping. First, the expression requirements to be rewritten as x^2+ax+bx-3. To uncover a and also b, collection up a mechanism to it is in solved.
Since abdominal is negative, a and also b have the the opposite signs. Because a+b is negative, the an adverse number has higher absolute value than the positive. The only such pair is the system solution.
Quadratic polynomial deserve to be factored making use of the revolution ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight), where x_1 and also x_2 room the options of the quadratic equation ax^2+bx+c=0.
All equations of the kind ax^2+bx+c=0 can be solved using the quadratic formula: frac-b±sqrtb^2-4ac2a. The quadratic formula provides two solutions, one when ± is addition and one once it is subtraction.

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Factor the initial expression making use of ax^2+bx+c=aleft(x-x_1 ight)left(x-x_2 ight). Substitute 3 because that x_1 and also -1 for x_2.
left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray together l together 2 & 0 & 3 \ -1 & 1 & 5 endarray ight>
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