Solved on Feb 04, 2024

Find the roots of a quadratic equation. Factor a polynomial by grouping.
Question 18: Find the solutions of $(x-4)(x-4)=0$ . Question 19: Factor $4 x^{3}+24 x+x^{2}+6$ by grouping.

STEP 1

Assumptions
1. For Question 18, we are solving the equation $(x-4)(x-4)=0$ .
2. For Question 19, we are factoring the polynomial $4x^3 + 24x + x^2 + 6$ by grouping.

STEP 2

To find the solutions for Question 18, we set each factor in the equation $(x-4)(x-4)=0$ equal to zero.
$x-4=0$

STEP 3

Solve for $x$ by adding 4 to both sides of the equation.
$x = 4$

STEP 4

Since both factors are the same, we only get one solution for $x$ .
The solution for Question 18 is $x=4$ .

STEP 5

For Question 19, we start by rewriting the polynomial in a way that makes grouping more apparent.
Rearrange the terms in the polynomial $4x^3 + 24x + x^2 + 6$ by ordering the terms by descending powers of $x$ .
$4x^3 + x^2 + 24x + 6$

STEP 6

Group the terms in pairs that have common factors.
$(4x^3 + x^2) + (24x + 6)$

STEP 7

Factor out the greatest common factor from each group.
For the first group, $x^2$ is the common factor:
$x^2(4x + 1)$
For the second group, $6$ is the common factor:
$6(4x + 1)$

STEP 8

Now that we have factored out the common factors, we can write the polynomial as:
$x^2(4x + 1) + 6(4x + 1)$

STEP 9

Notice that $(4x + 1)$ is a common factor in both terms. Factor out $(4x + 1)$ .
$(4x + 1)(x^2 + 6)$

STEP 10

The factored form of the polynomial $4x^3 + 24x + x^2 + 6$ is $(4x + 1)(x^2 + 6)$ .
The solutions for Question 18 and Question 19 are:
Question 18: $x=4$ (Answer D) Question 19: $(4x + 1)(x^2 + 6)$

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Find the roots of a quadratic equation. Factor a polynomial by grouping.
Question 18: Find the solutions of $(x-4)(x-4)=0$ . Question 19: Factor $4 x^{3}+24 x+x^{2}+6$ by grouping.

STEP 1

Assumptions
1. For Question 18, we are solving the equation $(x-4)(x-4)=0$ .
2. For Question 19, we are factoring the polynomial $4x^3 + 24x + x^2 + 6$ by grouping.

STEP 2

To find the solutions for Question 18, we set each factor in the equation $(x-4)(x-4)=0$ equal to zero.
$x-4=0$

STEP 3

Solve for $x$ by adding 4 to both sides of the equation.
$x = 4$

STEP 4

Since both factors are the same, we only get one solution for $x$ .
The solution for Question 18 is $x=4$ .

STEP 5

For Question 19, we start by rewriting the polynomial in a way that makes grouping more apparent.
Rearrange the terms in the polynomial $4x^3 + 24x + x^2 + 6$ by ordering the terms by descending powers of $x$ .
$4x^3 + x^2 + 24x + 6$

STEP 6

Group the terms in pairs that have common factors.
$(4x^3 + x^2) + (24x + 6)$

STEP 7

Factor out the greatest common factor from each group.
For the first group, $x^2$ is the common factor:
$x^2(4x + 1)$
For the second group, $6$ is the common factor:
$6(4x + 1)$

STEP 8

Now that we have factored out the common factors, we can write the polynomial as:
$x^2(4x + 1) + 6(4x + 1)$

STEP 9

Notice that $(4x + 1)$ is a common factor in both terms. Factor out $(4x + 1)$ .
$(4x + 1)(x^2 + 6)$

STEP 10

The factored form of the polynomial $4x^3 + 24x + x^2 + 6$ is $(4x + 1)(x^2 + 6)$ .
The solutions for Question 18 and Question 19 are:
Question 18: $x=4$ (Answer D) Question 19: $(4x + 1)(x^2 + 6)$

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Find the roots of a quadratic equation. Factor a polynomial by grouping.Question 18: Find the solutions of (x−4)(x−4)=0(x-4)(x-4)=0(x−4)(x−4)=0. Question 19: Factor 4x3+24x+x2+64 x^{3}+24 x+x^{2}+64x3+24x+x2+6 by grouping.

STEP 1

Assumptions1. For Question 18, we are solving the equation (x−4)(x−4)=0(x-4)(x-4)=0(x−4)(x−4)=0.2. For Question 19, we are factoring the polynomial 4x3+24x+x2+64x^3 + 24x + x^2 + 64x3+24x+x2+6 by grouping.

STEP 2

To find the solutions for Question 18, we set each factor in the equation (x−4)(x−4)=0(x-4)(x-4)=0(x−4)(x−4)=0 equal to zero.x−4=0x-4=0x−4=0

STEP 3

Solve for xxx by adding 4 to both sides of the equation.x=4x = 4x=4

STEP 4

Since both factors are the same, we only get one solution for xxx.The solution for Question 18 is x=4x=4x=4.

STEP 5

For Question 19, we start by rewriting the polynomial in a way that makes grouping more apparent.Rearrange the terms in the polynomial 4x3+24x+x2+64x^3 + 24x + x^2 + 64x3+24x+x2+6 by ordering the terms by descending powers of xxx.4x3+x2+24x+64x^3 + x^2 + 24x + 64x3+x2+24x+6

STEP 6

Group the terms in pairs that have common factors.(4x3+x2)+(24x+6)(4x^3 + x^2) + (24x + 6)(4x3+x2)+(24x+6)

STEP 7

Factor out the greatest common factor from each group.For the first group, x2x^2x2 is the common factor:x2(4x+1)x^2(4x + 1)x2(4x+1)For the second group, 666 is the common factor:6(4x+1)6(4x + 1)6(4x+1)

STEP 8

Now that we have factored out the common factors, we can write the polynomial as:x2(4x+1)+6(4x+1)x^2(4x + 1) + 6(4x + 1)x2(4x+1)+6(4x+1)

STEP 9

Notice that (4x+1)(4x + 1)(4x+1) is a common factor in both terms. Factor out (4x+1)(4x + 1)(4x+1).(4x+1)(x2+6)(4x + 1)(x^2 + 6)(4x+1)(x2+6)

STEP 10

The factored form of the polynomial 4x3+24x+x2+64x^3 + 24x + x^2 + 64x3+24x+x2+6 is (4x+1)(x2+6)(4x + 1)(x^2 + 6)(4x+1)(x2+6).The solutions for Question 18 and Question 19 are:Question 18: x=4x=4x=4 (Answer D) Question 19: (4x+1)(x2+6)(4x + 1)(x^2 + 6)(4x+1)(x2+6)

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Find the roots of a quadratic equation. Factor a polynomial by grouping.
Question 18: Find the solutions of $(x-4)(x-4)=0$ . Question 19: Factor $4 x^{3}+24 x+x^{2}+6$ by grouping.

Assumptions
1. For Question 18, we are solving the equation $(x-4)(x-4)=0$ .
2. For Question 19, we are factoring the polynomial $4x^3 + 24x + x^2 + 6$ by grouping.

To find the solutions for Question 18, we set each factor in the equation $(x-4)(x-4)=0$ equal to zero.
$x-4=0$

Solve for $x$ by adding 4 to both sides of the equation.
$x = 4$

Since both factors are the same, we only get one solution for $x$ .
The solution for Question 18 is $x=4$ .

For Question 19, we start by rewriting the polynomial in a way that makes grouping more apparent.
Rearrange the terms in the polynomial $4x^3 + 24x + x^2 + 6$ by ordering the terms by descending powers of $x$ .
$4x^3 + x^2 + 24x + 6$

Group the terms in pairs that have common factors.
$(4x^3 + x^2) + (24x + 6)$

Factor out the greatest common factor from each group.
For the first group, $x^2$ is the common factor:
$x^2(4x + 1)$
For the second group, $6$ is the common factor:
$6(4x + 1)$

Now that we have factored out the common factors, we can write the polynomial as:
$x^2(4x + 1) + 6(4x + 1)$

Notice that $(4x + 1)$ is a common factor in both terms. Factor out $(4x + 1)$ .
$(4x + 1)(x^2 + 6)$

The factored form of the polynomial $4x^3 + 24x + x^2 + 6$ is $(4x + 1)(x^2 + 6)$ .
The solutions for Question 18 and Question 19 are:
Question 18: $x=4$ (Answer D) Question 19: $(4x + 1)(x^2 + 6)$