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(a) Show that $(x - 1)$ is a factor of $f(x) = 2x^3 - 5x^2 + x + 2$ - Scottish Highers Maths - Question 2 - 2017

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(a)-Show-that-$(x---1)$-is-a-factor-of-$f(x)-=-2x^3---5x^2-+-x-+-2$-Scottish Highers Maths-Question 2-2017.png

(a) Show that $(x - 1)$ is a factor of $f(x) = 2x^3 - 5x^2 + x + 2$. (b) Hence, or otherwise, solve $f(x) = 0$.

Worked Solution & Example Answer:(a) Show that $(x - 1)$ is a factor of $f(x) = 2x^3 - 5x^2 + x + 2$ - Scottish Highers Maths - Question 2 - 2017

Step 1

(a) Show that $(x - 1)$ is a factor of $f(x) = 2x^3 - 5x^2 + x + 2$.

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Answer

To determine whether (x1)(x - 1) is a factor of the polynomial f(x)f(x), we can use the Factor Theorem. According to this theorem, (xc)(x - c) is a factor of f(x)f(x) if and only if f(c)=0f(c) = 0.

We will substitute x=1x = 1 into the polynomial:

f(1)=2(1)35(1)2+(1)+2f(1) = 2(1)^3 - 5(1)^2 + (1) + 2
=25+1+2= 2 - 5 + 1 + 2
=0= 0

Since f(1)=0f(1) = 0, it follows that (x1)(x - 1) is indeed a factor of f(x)f(x).

Step 2

(b) Hence, or otherwise, solve $f(x) = 0$.

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Answer

Now that we know (x1)(x - 1) is a factor, we can perform polynomial long division of f(x)f(x) by (x1)(x - 1) to find the other factors.

Dividing 2x35x2+x+22x^3 - 5x^2 + x + 2 by (x1)(x - 1):

  1. Divide the first term: 2x^3 ig/ x = 2x^2.
  2. Multiply (x1)(x - 1) by 2x22x^2: 2x32x22x^3 - 2x^2.
  3. Subtract: (5x2+2x2)+x+2=3x2+x+2( - 5x^2 + 2x^2) + x + 2 = -3x^2 + x + 2.
  4. Repeat: Divide -3x^2 ig/ x = -3x; multiply and subtract.
  5. Finally, this gives 22.

Thus, we get the factorization:

f(x)=(x1)(2x23x2)f(x) = (x - 1)(2x^2 - 3x - 2)

Now, we can factor 2x23x22x^2 - 3x - 2 further:

2x23x2=(2x+1)(x2)2x^2 - 3x - 2 = (2x + 1)(x - 2)

Thus, f(x)f(x) can be factored as:

f(x)=(x1)(2x+1)(x2)f(x) = (x - 1)(2x + 1)(x - 2)

Setting f(x)=0f(x) = 0 gives:

  1. x1=0ightarrowx=1x - 1 = 0 ightarrow x = 1.
  2. 2x + 1 = 0 ightarrow x = - rac{1}{2}.
  3. x2=0ightarrowx=2x - 2 = 0 ightarrow x = 2.

Therefore, the solutions to the equation f(x)=0f(x) = 0 are:

x = 1, \, x = - rac{1}{2}, \, x = 2.

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