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3 years ago

Simplify the expression (4x − 3)(x + 5). 4x2 − 17x + 15 4x2 − 17x − 15 4x2 + 17x + 15 4x2 + 17x − 15

2 answers:

8 0

A) $(4x-3)(x+5)=$

$*(a+b)(c+d)=ac+ad+bc+bd$

$Soo..$

$4x*x+4x*5-3*x-3*5=$

$=4 x^{2} +20x-3x-15$

$=4x^{2} +17x-15$

b) $4x^{2} -17x+15=$

$=(4x-5)(x-3)$

c) $4 x^{2} -17x-15=$

$=(4x+3)(x-5)$

d) $4 x^{2} +17x+15=$
$=(4x+5)(x+3)$

e) $4 x^{2} +17x-15=$
$=(4x-3)(x+5)$

DanielleElmas [232]3 years ago

5 0

There are mnemonic acronyms out there for reminding you how to do this. I think of it simply as using the distributive property.
(4x-3)(x+5) = 4x(x+5) -3(x+5)
then again
= (4x² +20x) + (-3x -15)
= 4x² + (20-3)x -15 . . . . . . . combine like terms
= 4x² +17x -15

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Write the polynomial in factored form. <br><br> p(x)=(x+5)(x-___)(x+___)

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Answer:

Step-by-step explanation:

Given : $p(x)=x^{3} +6x^{2} -7x-60$

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Part A:

First find the potential roots of p(x) using rational root theorem;

So, $\text{Possible roots} =\pm\frac{\text{factors of constant term}}{\text{factors of leading coefficient}}$

Since constant term = -60

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$\text{Possible roots} =\pm\frac{1,2,3,4,5,6,10,12,15,20,60}{1}$

Thus the possible roots are $\pm1, \pm2, \pm 3, \pm4, \pm5,\pm6, \pm10, \pm12, \pm15, \pm20, \pm60$

Thus from the given options the correct answers are -10,-5,3,15

Now For Part B we will use synthetic division

Out of the possible roots we will use the root which gives remainder 0 in synthetic division :

Since we can see in the figure With -5 we are getting 0 remainder.

Refer the attached figure

We have completed the table and have obtained the following resulting coefficients: 1 , 1,−12,0. All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.

Thus the quotient is $x^{2} +x-12$