How do you multiply $(4 x - 5) (2 x^{5} + x 3 - 1)$ $(4x-5)(2x^5 + x3 - 1)$ ?

Question

How do you multiply $(4 x - 5) (2 x^{5} + x 3 - 1)$ $(4x-5)(2x^5 + x3 - 1)$ ?

1 Answer

Impact of this question

1708 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-26T19:19:22

For each power of $x$ $x$ from $x^{6}$ $x^6$ down to $x^{0}$ $x^0$ , pick out the pairs that multiply to give that power and add them to find:

$(4 x - 5) (2 x^{5} + x^{3} - 1)$ $(4x-5)(2x^5+x^3-1)$

$= 8 x^{6} - 10 x^{5} + 4 x^{4} - 5 x^{3} - 4 x + 5$ $= 8x^6-10x^5+4x^4-5x^3-4x+5$

Explanation:

Looking at each power of $x$ $x$ in turn from $x^{6}$ $x^6$ down to $x^{0}$ $x^0$ , pick out the pairs that multiply to give a term with that power of $x$ $x$ :

$x^{6}$ $x^6$ : $4 x \cdot 2 x^{5} = 8 x^{6}$ $4x*2x^5 = 8x^6$

$x^{5}$ $x^5$ : $- 5 \cdot 2 x^{5} = - 10 x^{5}$ $-5*2x^5 = -10x^5$

$x^{4}$ $x^4$ : $4 x \cdot x^{3} = 4 x^{4}$ $4x*x^3 = 4x^4$

$x^{3}$ $x^3$ : $- 5 \cdot x^{3} = - 5 x^{3}$ $-5 * x^3 = -5x^3$

$x^{2}$ $x^2$ : none

$x$ $x$ : $4 x \cdot - 1 = - 4 x$ $4x*-1 = -4x$

$1$ $1$ : $- 5 \cdot - 1 = 5$ $-5*-1 = 5$

Add to get:

$8 x^{6} - 10 x^{5} + 4 x^{4} - 5 x^{3} - 4 x + 5$ $8x^6-10x^5+4x^4-5x^3-4x+5$

Normally when multiplying two polynomials in this way you would have two pairs to multiply and add for most of the powers of $x$ $x$ , but a similar approach works.

Science

Math

Humanities

... and beyond

How do you multiply $(4 x - 5) (2 x^{5} + x 3 - 1)$ $(4x-5)(2x^5 + x3 - 1)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you multiply (4x-5)(2x^5 + x3 - 1)(4x−5)(2x5+x3−1)(4x-5)(2x^5 + x3 - 1)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you multiply $(4 x - 5) (2 x^{5} + x 3 - 1)$ $(4x-5)(2x^5 + x3 - 1)$ ?