Mastery Check Solutions

Show What You Know

A student was asked to show their work to prove the equation represented a polynomial identity. Find their error and correct it to prove the identity exists. Indicate the line(s) in which the error occurs.
$x^{4} - y^{4} = (x^{2} + y^{2}) (x + y) (x - y)$

Simplifying the right side

$(x^{2} + y^{2}) (x + y) (x - y) x^{3} + x^{2} y + x^{3} - x^{2} y$
$+ x y^{2} + y^{3} + x y^{2} - y^{3}$
$2 x^{3} + 2 x y^{2}$

Student Response:

I distributed the 1st binomial across the 2nd and 3rd binomial.

Note

You should indicate the errors are located in the highlighted expression and words.

$x^{4} - y^{4} \neq 2 x^{3} + 2 x y^{2}$

This is NOT an identity because the left and right sides of the equation are NOT equal.

Corrected Right side (sample)

$(x^{2} + y^{2}) (x^{2} - x y + x y - y^{2})$

The 2nd and 3rd binomials form a difference of two squares identity when multiplied.

$(x^{2} + y^{2}) (x^{2} - y^{2})$

Now the 1st binomial and the product of the 2nd and 3rd are multiplied together.

$x^{4} - x^{2} y^{2} + x^{2} y^{2} - y^{4} x^{4} - y^{4}$

When simplified, the left and right sides are equal, forming an identity.

Note

You can multiply any two binomials together. Once you have that product, you can multiply it by the 3rd binomial in the expression. It is more efficient to multiply the 2nd and 3rd binomials because they represent the difference of two squares.

Determine the non-zero value of Q that will form a polynomial identity. Then rewrite the polynomial using the value you found for Q.
${(Q x + 1)}^{2} = Q x^{2} + (Q x + 2) (5 x + 1) - 4 x - 1$

Left side ${(Q x + 1)}^{2}$	Right side $Q x^{2} + (Q x + 2) (5 x + 1) - 4 x - 1$
$Q^{2} x^{2} + 2 Q x + 1$	$Q x^{2} + 5 Q x^{2} + Q x + 10 x + 2 - 4 x - 1$
	$6 Q x^{2} + Q x + 6 x + 1$
Compare linear terms: $2 Q x = Q x + 6 x 2 Q = Q + 6 - Q - Q Q = 6$	The polynomial identity is: ${(6 x + 1)}^{2} = 6 x^{2} + (6 x + 2) (5 x + 1) - 4 x - 1$

Note

Any terms of the same degree can be compared to find the value of Q. Using the linear terms may be more efficient in this case because factoring is not needed to find the missing value.
Using the degree 2 terms:
$\begin{array}{rcl} Q^{2} x^{2} & = & Q x^{2} + 5 Q x^{2} \\ Q^{2} & = & Q + 5 Q \\ Q^{2} - 6 Q & = & 0 \\ Q (Q - 6) & = & 0 \\ Q = 0, Q & = & 6 \end{array}$

Say What You Know

In your own words, talk about what you have learned using the objectives for this part of the lesson and your work on this page.

Note

Restate the objectives of the lesson in your own words. If you are unable to restate the lesson objectives, go back and reread the objectives and then explain them.

Determine if a polynomial identity exists.
Determine the value of an unknown to make a polynomial expression or equation true.