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Sagot :
Certainly! Here's a detailed step-by-step solution to the problem:
### Given:
1. [tex]\( (5x - 1) \)[/tex] is a factor of [tex]\( 5x^3 + ax^2 + bx + 6 \)[/tex].
2. The remainder when the polynomial is divided by [tex]\( (x - 2) \)[/tex] is 72.
#### Part (a): Find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
Since [tex]\((5x - 1)\)[/tex] is a factor of [tex]\(5x^3 + ax^2 + bx + 6\)[/tex], it implies that [tex]\(P(\frac{1}{5}) = 0\)[/tex].
Given polynomial [tex]\(P(x) = 5x^3 + ax^2 + bx + 6\)[/tex], substituting [tex]\(\frac{1}{5}\)[/tex] into [tex]\(P(x)\)[/tex]:
[tex]\[ P\left(\frac{1}{5}\right) = 5\left(\frac{1}{5}\right)^3 + a\left(\frac{1}{5}\right)^2 + b\left(\frac{1}{5}\right) + 6 = 0 \][/tex]
Simplify:
[tex]\[ 5 \left(\frac{1}{125}\right) + a \left(\frac{1}{25}\right) + b \left(\frac{1}{5}\right) + 6 = 0 \][/tex]
[tex]\[ \frac{5}{125} + \frac{a}{25} + \frac{b}{5} + 6 = 0 \][/tex]
[tex]\[ \frac{1}{25} + \frac{a}{25} + \frac{5b}{25} + 6 = 0 \][/tex]
[tex]\[ \frac{1 + a + 5b}{25} + 6 = 0 \][/tex]
[tex]\[ 1 + a + 5b + 150 = 0 \][/tex]
[tex]\[ a + 5b + 151 = 0 \][/tex]
[tex]\[ a + 5b = -151 \][/tex] ... (1)
Also, the remainder when divided by [tex]\((x - 2)\)[/tex] is 72, implying [tex]\(P(2) = 72\)[/tex]:
[tex]\[ P(2) = 5(2)^3 + a(2)^2 + b(2) + 6 = 72 \][/tex]
[tex]\[ 40 + 4a + 2b + 6 = 72 \][/tex]
[tex]\[ 46 + 4a + 2b = 72 \][/tex]
[tex]\[ 4a + 2b = 26 \][/tex]
[tex]\[ 2a + b = 13 \][/tex] ... (2)
Using equations (1) and (2), solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
1. [tex]\(a + 5b = -151\)[/tex]
2. [tex]\(2a + b = 13\)[/tex]
From equation (2):
[tex]\[ b = 13 - 2a \][/tex]
Substitute [tex]\(b\)[/tex] into equation (1):
[tex]\[ a + 5(13 - 2a) = -151 \][/tex]
[tex]\[ a + 65 - 10a = -151 \][/tex]
[tex]\[ -9a = -216 \][/tex]
[tex]\[ a = 24 \][/tex]
Substituting [tex]\(a = 24\)[/tex] back into equation (2):
[tex]\[ 2(24) + b = 13 \][/tex]
[tex]\[ 48 + b = 13 \][/tex]
[tex]\[ b = -35 \][/tex]
So, the values are:
[tex]\[ a = 24 \][/tex]
[tex]\[ b = -35 \][/tex]
#### Part (b): Factorize the expression completely
Now substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex] back into the polynomial:
[tex]\[ 5x^3 + 24x^2 - 35x + 6 \][/tex]
Given that [tex]\( (5x - 1) \)[/tex] is a factor, let's divide the polynomial by [tex]\( (5x - 1) \)[/tex]:
To perform polynomial division:
1. [tex]\(5x^3 + 24x^2 - 35x + 6\)[/tex] divided by [tex]\(5x - 1\)[/tex]:
Perform the division (performing synthetic or long division):
[tex]\[ \frac{5x^3 + 24x^2 - 35x + 6}{5x - 1} = x^2 + 5x - 6 \][/tex]
So, the polynomial can be factorized as:
[tex]\[ 5x^3 + 24x^2 - 35x + 6 = (5x - 1)(x^2 + 5x - 6) \][/tex]
Factorize [tex]\(x^2 + 5x - 6\)[/tex]:
[tex]\[ x^2 + 5x - 6 = (x + 6)(x - 1) \][/tex]
So, the complete factorization is:
[tex]\[ 5x^3 + 24x^2 - 35x + 6 = (5x - 1)(x + 6)(x - 1) \][/tex]
#### Part (c): Solve the equation [tex]\(5x^3 + 24x^2 - 35x + 6 = 0\)[/tex]
Using the factorized form:
[tex]\[ (5x - 1)(x + 6)(x - 1) = 0 \][/tex]
Set each factor to zero:
1. [tex]\(5x - 1 = 0 \Rightarrow x = \frac{1}{5}\)[/tex]
2. [tex]\(x + 6 = 0 \Rightarrow x = -6\)[/tex]
3. [tex]\(x - 1 = 0 \Rightarrow x = 1\)[/tex]
So, the solutions are:
[tex]\[ x = \frac{1}{5}, x = -6, x = 1 \][/tex]
Therefore, the roots of the polynomial [tex]\(5x^3 + 24x^2 - 35x + 6 = 0\)[/tex] are:
[tex]\[ \boxed{\frac{1}{5}, -6, 1} \][/tex]
### Given:
1. [tex]\( (5x - 1) \)[/tex] is a factor of [tex]\( 5x^3 + ax^2 + bx + 6 \)[/tex].
2. The remainder when the polynomial is divided by [tex]\( (x - 2) \)[/tex] is 72.
#### Part (a): Find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
Since [tex]\((5x - 1)\)[/tex] is a factor of [tex]\(5x^3 + ax^2 + bx + 6\)[/tex], it implies that [tex]\(P(\frac{1}{5}) = 0\)[/tex].
Given polynomial [tex]\(P(x) = 5x^3 + ax^2 + bx + 6\)[/tex], substituting [tex]\(\frac{1}{5}\)[/tex] into [tex]\(P(x)\)[/tex]:
[tex]\[ P\left(\frac{1}{5}\right) = 5\left(\frac{1}{5}\right)^3 + a\left(\frac{1}{5}\right)^2 + b\left(\frac{1}{5}\right) + 6 = 0 \][/tex]
Simplify:
[tex]\[ 5 \left(\frac{1}{125}\right) + a \left(\frac{1}{25}\right) + b \left(\frac{1}{5}\right) + 6 = 0 \][/tex]
[tex]\[ \frac{5}{125} + \frac{a}{25} + \frac{b}{5} + 6 = 0 \][/tex]
[tex]\[ \frac{1}{25} + \frac{a}{25} + \frac{5b}{25} + 6 = 0 \][/tex]
[tex]\[ \frac{1 + a + 5b}{25} + 6 = 0 \][/tex]
[tex]\[ 1 + a + 5b + 150 = 0 \][/tex]
[tex]\[ a + 5b + 151 = 0 \][/tex]
[tex]\[ a + 5b = -151 \][/tex] ... (1)
Also, the remainder when divided by [tex]\((x - 2)\)[/tex] is 72, implying [tex]\(P(2) = 72\)[/tex]:
[tex]\[ P(2) = 5(2)^3 + a(2)^2 + b(2) + 6 = 72 \][/tex]
[tex]\[ 40 + 4a + 2b + 6 = 72 \][/tex]
[tex]\[ 46 + 4a + 2b = 72 \][/tex]
[tex]\[ 4a + 2b = 26 \][/tex]
[tex]\[ 2a + b = 13 \][/tex] ... (2)
Using equations (1) and (2), solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
1. [tex]\(a + 5b = -151\)[/tex]
2. [tex]\(2a + b = 13\)[/tex]
From equation (2):
[tex]\[ b = 13 - 2a \][/tex]
Substitute [tex]\(b\)[/tex] into equation (1):
[tex]\[ a + 5(13 - 2a) = -151 \][/tex]
[tex]\[ a + 65 - 10a = -151 \][/tex]
[tex]\[ -9a = -216 \][/tex]
[tex]\[ a = 24 \][/tex]
Substituting [tex]\(a = 24\)[/tex] back into equation (2):
[tex]\[ 2(24) + b = 13 \][/tex]
[tex]\[ 48 + b = 13 \][/tex]
[tex]\[ b = -35 \][/tex]
So, the values are:
[tex]\[ a = 24 \][/tex]
[tex]\[ b = -35 \][/tex]
#### Part (b): Factorize the expression completely
Now substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex] back into the polynomial:
[tex]\[ 5x^3 + 24x^2 - 35x + 6 \][/tex]
Given that [tex]\( (5x - 1) \)[/tex] is a factor, let's divide the polynomial by [tex]\( (5x - 1) \)[/tex]:
To perform polynomial division:
1. [tex]\(5x^3 + 24x^2 - 35x + 6\)[/tex] divided by [tex]\(5x - 1\)[/tex]:
Perform the division (performing synthetic or long division):
[tex]\[ \frac{5x^3 + 24x^2 - 35x + 6}{5x - 1} = x^2 + 5x - 6 \][/tex]
So, the polynomial can be factorized as:
[tex]\[ 5x^3 + 24x^2 - 35x + 6 = (5x - 1)(x^2 + 5x - 6) \][/tex]
Factorize [tex]\(x^2 + 5x - 6\)[/tex]:
[tex]\[ x^2 + 5x - 6 = (x + 6)(x - 1) \][/tex]
So, the complete factorization is:
[tex]\[ 5x^3 + 24x^2 - 35x + 6 = (5x - 1)(x + 6)(x - 1) \][/tex]
#### Part (c): Solve the equation [tex]\(5x^3 + 24x^2 - 35x + 6 = 0\)[/tex]
Using the factorized form:
[tex]\[ (5x - 1)(x + 6)(x - 1) = 0 \][/tex]
Set each factor to zero:
1. [tex]\(5x - 1 = 0 \Rightarrow x = \frac{1}{5}\)[/tex]
2. [tex]\(x + 6 = 0 \Rightarrow x = -6\)[/tex]
3. [tex]\(x - 1 = 0 \Rightarrow x = 1\)[/tex]
So, the solutions are:
[tex]\[ x = \frac{1}{5}, x = -6, x = 1 \][/tex]
Therefore, the roots of the polynomial [tex]\(5x^3 + 24x^2 - 35x + 6 = 0\)[/tex] are:
[tex]\[ \boxed{\frac{1}{5}, -6, 1} \][/tex]
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