element the expression through grouping. First, the expression demands to it is in rewritten as x^2+ax+bx+16. To find a and b, collection up a mechanism to be solved.

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Since abdominal muscle is positive, a and also b have actually the same sign. Due to the fact that a+b is negative, a and also b are both negative. Perform all such integer pairs that give product 16.
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This is a perfect square trinomial, since the first and last terms room perfect squaresdisplaystyleleft(sqrtx^2=xquad extandquadsqrt16=4 ight) ...
x2-8x+10 Final an outcome : x2 - 8x + 10 action by action solution : action 1 :Trying to element by dividing the center term 1.1 Factoring x2-8x+10 The very first term is, x2 its coefficient is 1 . ...
What is the remainder when x^100 -8x^99+12x^98-3x^10+24x^9-36x^8+3x^2-29x + 41 is separated by: x^2-8x+12.
https://math.stackexchange.com/questions/313044/what-is-the-remainder-when-x100-8x9912x98-3x1024x9-36x83x
very first note that r(x)=ax+b for part a,binmathbb Z for x=2 you get r(2)=-5 (not r(x)=-5) Rightarrow 2a+b=-5. Carry out the same for x=6 to discover r(6).Then uncover a,b.
displaystylex^2-8x+15=left(x-5 ight)cdotleft(x-3 ight) Explanation: displaystylex^2-8x+15=left(x-5 ight)cdotleft(x-3 ight)
x2-8x+17 Final an outcome : x2 - 8x + 17 action by action solution : action 1 :Trying to variable by separating the middle term 1.1 Factoring x2-8x+17 The very first term is, x2 its coefficient is 1 . ...
-3x2-8x+16 Final an outcome : (4 - 3x) • (x + 4) action by action solution : action 1 :Equation at the finish of step 1 : ((0 - 3x2) - 8x) + 16 step 2 : step 3 :Pulling out prefer terms : 3.1 pull ...
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Factor the expression by grouping. First, the expression requirements to be rewritten as x^2+ax+bx+16. To discover a and also b, set up a device to be solved.
Since abdominal is positive, a and b have the exact same sign. Due to the fact that a+b is negative, a and b are both negative. Perform all together integer bag that offer product 16.

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This trinomial has the type of a trinomial square, possibly multiplied by a usual factor. Trinomial squares deserve to be factored by recognize the square root of the leading and trailing terms.