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Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.

Daryl factors the polynomial p(x)=x3+x2−26x+24 to rewrite it as p(x)=(x+6)(x−4)(x−1). which equations must be true? there may be more than one correct answer. select all correct answers. p(1)=0 p(−4)=0 p(−1)=0 p(6)=0 p(4)=0 p(−6)=0

How do i put 3(x+7) in distributive property