It is given that $f(x)=x^3+ax^2+bx+5$ , where $a$ and $b$ are constants, has a factor of $x-1$ and leaves a remainder of $-1$ when divided by $x-2$ .

(a)
Find the values of $a$ and $b$ .
(b)
Using these values of $a$ and $b$ , factorise $f(x)$ completely into three linear factors, using surds where necessary.

Remainder factor theorems Simultaneous one linear Operations rationalising 中文

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Worked solution

3(b)

1. Substitute the values of $a$ and $b$ from part (a) into $f(x)$

From part (a), $a = -1$ and $b = -5$ .
$f(x) = x^3 - x^2 - 5x + 5$

2. Factorise out the known linear factor $(x-1)$

Since $(x-1)$ is a factor, we can factorise by grouping or long division:
$x^3 - x^2 - 5x + 5 = x^2(x-1) - 5(x-1)$
$= (x-1)(x^2 - 5)$

3. Factorise the quadratic factor completely using surds

Using the difference of two squares $a^2 - b^2 = (a-b)(a+b)$ :
$x^2 - 5 = (x - \sqrt{5})(x + \sqrt{5})$
Therefore, $f(x) = (x-1)(x-\sqrt{5})(x+\sqrt{5})$

Answer

$(x-1)(x-\sqrt{5})(x+\sqrt{5})$

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It is given that f(x)=x3+ax2+bx+5f(x)=x^3+ax^2+bx+5f(x)=x3+ax2+bx+5, where aaa and bbb are constants, has a factor of x−1x-1x−1 and leaves a remainder of −1-1−1 when divided by x−2x-2x−2.

Worked solution

3(b)

More questions on this topic

It is given that $f(x)=x^3+ax^2+bx+5$ , where $a$ and $b$ are constants, has a factor of $x-1$ and leaves a remainder of $-1$ when divided by $x-2$ .