The expression (y2−t2)(y+k) can be written as y3+36y2−9y+s where t,k, and s are constants. What is the value of s ?Choose 1 answer:(A) −324(B) −54(C) 54(D) 324
Q. The expression (y2−t2)(y+k) can be written as y3+36y2−9y+s where t,k, and s are constants. What is the value of s ?Choose 1 answer:(A) −324(B) −54(C) 54(D) 324
Factorize Difference of Squares: We need to expand the expression y2−t2)(y+k)andcompareitto$y3+36y2−9y+s to find the value of s.
Expand and Compare Expressions: First, recognize that y2−t2 is a difference of squares and can be factored into (y+t)(y−t).
Identify Coefficients: Now, distribute (y+t)(y−t) across (y+k) to get the expanded form.(y+t)(y−t)(y+k)=y(y2−t2)+k(y2−t2)
Eliminate Terms: Expand y(y2−t2) to get y3−yt2.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2.Now, compare the expanded expression to y3+36y2−9y+s. We see that the coefficient of y2 in the expanded expression is k−t, which must equal 36 from the given expression.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2.Now, compare the expanded expression to y3+36y2−9y+s. We see that the coefficient of y2 in the expanded expression is k−t, which must equal 36 from the given expression.The term −yt2 has no corresponding term in y3+36y2−9y+s, which means that t must be ky2−kt20 to eliminate the term ky2−kt21.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2.Now, compare the expanded expression to y3+36y2−9y+s. We see that the coefficient of y2 in the expanded expression is k−t, which must equal 36 from the given expression.The term −yt2 has no corresponding term in y3+36y2−9y+s, which means that t must be ky2−kt20 to eliminate the term ky2−kt21.With t being ky2−kt20, the term ky2−kt24 also disappears, leaving us with ky2−kt25 as the only terms with y2 and ky2−kt27.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2. Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2. Now, compare the expanded expression to y3+36y2−9y+s. We see that the coefficient of y2 in the expanded expression is k−t, which must equal 36 from the given expression. The term −yt2 has no corresponding term in y3+36y2−9y+s, which means that t must be ky2−kt20 to eliminate the term ky2−kt21. With t being ky2−kt20, the term ky2−kt24 also disappears, leaving us with ky2−kt25 as the only terms with y2 and ky2−kt27. Since there is no ky2−kt27 term in y3+36y2−9y+s, y3−yt2+ky2−kt20 must also be ky2−kt20 to eliminate the term y3−yt2+ky2−kt22.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2.Now, compare the expanded expression to y3+36y2−9y+s. We see that the coefficient of y2 in the expanded expression is k−t, which must equal 36 from the given expression.The term −yt2 has no corresponding term in y3+36y2−9y+s, which means that t must be ky2−kt20 to eliminate the term ky2−kt21.With t being ky2−kt20, the term ky2−kt24 also disappears, leaving us with ky2−kt25 as the only terms with y2 and ky2−kt27.Since there is no ky2−kt27 term in y3+36y2−9y+s, y3−yt2+ky2−kt20 must also be ky2−kt20 to eliminate the term y3−yt2+ky2−kt22.Now we know that y3−yt2+ky2−kt23 and y3−yt2+ky2−kt24, so the term y3−yt2+ky2−kt25 in the expression y3+36y2−9y+s is the constant term left over, which is ky2−kt24.
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2.Now, compare the expanded expression to y3+36y2−9y+s. We see that the coefficient of y2 in the expanded expression is k−t, which must equal 36 from the given expression.The term −yt2 has no corresponding term in y3+36y2−9y+s, which means that t must be ky2−kt20 to eliminate the term ky2−kt21.With t being ky2−kt20, the term ky2−kt24 also disappears, leaving us with ky2−kt25 as the only terms with y2 and ky2−kt27.Since there is no ky2−kt27 term in y3+36y2−9y+s, y3−yt2+ky2−kt20 must also be ky2−kt20 to eliminate the term y3−yt2+ky2−kt22.Now we know that y3−yt2+ky2−kt23 and y3−yt2+ky2−kt24, so the term y3−yt2+ky2−kt25 in the expression y3+36y2−9y+s is the constant term left over, which is ky2−kt24.Substitute y3−yt2+ky2−kt23 and y3−yt2+ky2−kt24 into ky2−kt24 to find y3−yt2+ky2−kt25.y3+36y2−9y+s2y3+36y2−9y+s3
Find Constant Term: Expand k(y2−t2) to get ky2−kt2.Combine the terms to get the full expanded expression: y3−yt2+ky2−kt2.Now, compare the expanded expression to y3+36y2−9y+s. We see that the coefficient of y2 in the expanded expression is k−t, which must equal 36 from the given expression.The term −yt2 has no corresponding term in y3+36y2−9y+s, which means that t must be ky2−kt20 to eliminate the term ky2−kt21.With t being ky2−kt20, the term ky2−kt24 also disappears, leaving us with ky2−kt25 as the only terms with y2 and ky2−kt27.Since there is no ky2−kt27 term in y3+36y2−9y+s, y3−yt2+ky2−kt20 must also be ky2−kt20 to eliminate the term y3−yt2+ky2−kt22.Now we know that y3−yt2+ky2−kt23 and y3−yt2+ky2−kt24, so the term y3−yt2+ky2−kt25 in the expression y3+36y2−9y+s is the constant term left over, which is ky2−kt24.Substitute y3−yt2+ky2−kt23 and y3−yt2+ky2−kt24 into ky2−kt24 to find y3−yt2+ky2−kt25. y3+36y2−9y+s2y3+36y2−9y+s3However, we made a mistake. We should have compared the constant terms directly instead of assuming t and y3−yt2+ky2−kt20 are zero. Let's correct this by comparing the constant term ky2−kt24 from the expanded expression to the constant term y3−yt2+ky2−kt25 in the given expression y3+36y2−9y+s.
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