SAT Math - How to multiply polynomials

$F(x) = x^{3} + x^{2} - x + 2$

and

$G(x) = x^{2} + 5$

What is ?

$(FG)(x) = x^{5} + x^{4} +4x^{3} + 7x^{2} - 5x +10$

$(FG)(x) = x^{5} + x^{4} - x^{3} + 2x^{2} - 5x -10$

$(FG)(x) = x^{5} + x^{4} - x - 2$

$(FG)(x) = x^{3} + 2x^{2} - x + 7$

$(FG)(x) = x^{3} - x - 3$

Explanation

$(FG)(x) = F(x)G(x)$ so we multiply the two function to get the answer. We use $x^{m}x^{n} = x^{m+n}$

represents a positive quantity; represents a negative quantity.

$x^{4} = 116$

$y^{2} = 11$

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

The correct answer is not among the other choices.

Explanation

The first two binomials are the difference and the sum of the same two expressions, which, when multiplied, yield the difference of their squares:

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

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

Again, a sum is multiplied by a difference to yield a difference of squares, which by the Power of a Power Property, is equal to:

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

$= x^{2\cdot 2} - y^{2\cdot 2}$

$= x^{4} - y^{4}$

$y^{2} = 11$ , so by the Power of a Power Property,

$(y^{2} )^{2}= 11^{2}$

$y^{2\cdot 2} = 121$

$y^{4} = 121$

Also, $x^{4} = 116$ , so we can now substitute accordingly:

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

$= x^{4} - y^{4}$

Note that the signs of and are actually irrelevant to the problem.

represents a positive quantity; represents a negative quantity.

$x^{6} = 625$

$y ^{6} = 64$

Evaluate $(x-y) (x^{2}+xy+y^{2})$ .

Explanation

$(x-y) (x^{2}+xy+y^{2})$ can be recognized as the pattern conforming to that of the difference of two perfect cubes:

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

Additionally, by way of the Power of a Power Property,

$(x^{3})^{ 2 } = x^{3 \cdot 2} = x^{6 }$ , making $x^{3}$ a square root of $x^{6}$ , or 625; since is positive, so is $x^{3}$ , so

$x^{3} = \sqrt{625} = 25$ .

Similarly, $y ^{3}$ is a square root of $y ^{6}$ , or 64; since is negative, so is $y^{3}$ (as an odd power of a negative number is negative), so

$y^{3} = - \sqrt{64} = -8$ .

Therefore, substituting:

$(x-y) (x^{2}+xy+y^{2}) = x^{3} - y^{3} = 25 - (-8) = 25+8= 33$ .

and represent positive quantities.

$x^{2} = 20$

$y^{2} = 5$

Evaluate $(x-y) (x^{2}+xy+y^{2})$ .

$35 \sqrt{5}$

$45\sqrt{5}$

$5 \sqrt{5}$

$35 \sqrt{15 }$

$45 \sqrt{15 }$

Explanation

$(x-y) (x^{2}+xy+y^{2})$ can be recognized as the pattern conforming to that of the difference of two perfect cubes:

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

Additionally,

$x^{2} = 20$ and is positive, so

$x = \sqrt{20}$

Using the product of radicals property, we see that

$x = \sqrt{4 \cdot 5} = \sqrt{4 } \cdot \sqrt{ 5} = 2 \sqrt{ 5}$

and

$x ^{3 } = x^{2} \cdot x = 20 \cdot 2 \sqrt{5} = 40 \sqrt{5}$

$y^{2} = 5$ and is positive, so

$y = \sqrt{5}$ ,

and

$y^{3}= y^{2} \cdot y = 5 \sqrt{5}$

Substituting for $x ^{3 }$ and $y^{3}$ , then collecting the like radicals,

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

$= x^{3} - y^{3}$

$= 40 \sqrt{5}-5 \sqrt{5}$

$=( 40 -5 )\sqrt{5}$

$=35 \sqrt{5}$ .

Find the product:

Explanation

Find the product:

Step 1: Use the distributive property.

Step 2: Combine like terms.

How to multiply polynomials

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