Question Video: Finding the Value of a Constant Using Division by a Monomial | Nagwa Question Video: Finding the Value of a Constant Using Division by a Monomial | Nagwa

Question Video: Finding the Value of a Constant Using Division by a Monomial Mathematics • First Year of Preparatory School

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Find the value of the constant 𝑘 given that (6𝑥⁵ + 𝑘𝑥⁷)/2𝑥² = 3𝑥³ + 4𝑥⁵.

03:44

Video Transcript

Find the value of the constant 𝑘 given that six 𝑥 raised to the fifth power plus 𝑘𝑥 raised to the seventh power all over two 𝑥 squared is equal to three 𝑥 cubed plus four 𝑥 raised to the fifth power.

In this question, we want to find the value of a constant 𝑘 by using a given equation involving operations on polynomials. There are many ways of answering this question, and we will go through two of these. The first way that we can find the value of 𝑘 is to note that the left-hand side of the equation is the quotient of a polynomial and a monomial. We can divide a polynomial by a monomial by dividing each of the terms by the monomial separately and applying the quotient rule for exponents. First, we split the division over each term to obtain the following equation. Next, we recall that the quotient rule for exponents tells us that if 𝑥 is nonzero, then 𝑥 raised to the power of 𝑚 over 𝑥 raised to the power of 𝑛 is equal to 𝑥 raised to the power of 𝑚 minus 𝑛.

It is worth noting that we know that 𝑥 is nonzero since we cannot divide by zero. We can apply this to each term on the left-hand side of the equation. First, we have six over two is three. And we need to multiply this by 𝑥 raised to the power of the difference in the exponents, that is, five minus two. We can follow the same process for the second term to get 𝑘 over two multiplied by 𝑥 raised to the power of seven minus two. This expression must be equal to three 𝑥 cubed plus four 𝑥 raised to the fifth power. We can then evaluate the exponents to obtain the following equation.

It is worth reiterating at this point that we are not working with an equation in the traditional sense. This is an equivalence. We want both sides of the equation to be equal for all valid values of 𝑥, whereas in an equation, we are looking for the values of a variable that satisfy the equation. We can now note that the first terms of both expressions are identical. Similarly, the second term in both expressions has the same variable factor of 𝑥 raised to the fifth power. Therefore, for the expressions to be equivalent, the coefficients of the terms must be equal. So, 𝑘 over two must be equal to four, which we can calculate is true only when 𝑘 equals eight.

It is worth noting that we can answer this question without using division by instead multiplying the equation through by two 𝑥 squared. Doing this gives us the following equivalence, which we need to solve for 𝑘. We can simplify the expression on the right-hand side by distributing the monomial factor and multiplying the powers of 𝑥 by adding their exponents. We can then evaluate the expressions in the exponents and once again note that the expressions are identical, so we must have that 𝑘 equals eight.

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