Solve 8a ^ 4 + 2a ^ 2 - 45 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$8a^{4}+2a^{2}-45$$

Factor

$\left(2a-3\right)\left(2a+3\right)\left(2a^{2}+5\right)$

Solution Steps

To factor the expression, solve the equation where it equals to $0$.

$$8a^{4}+2a^{2}-45=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-45$ and $q$ divides the leading coefficient $8$. List all candidates $\frac{p}{q}$.

$$±\frac{45}{8},±\frac{45}{4},±\frac{45}{2},±45,±\frac{15}{8},±\frac{15}{4},±\frac{15}{2},±15,±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{5}{8},±\frac{5}{4},±\frac{5}{2},±5,±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$a=\frac{3}{2}$$

By Factor theorem, $a-k$ is a factor of the polynomial for each root $k$. Divide $8a^{4}+2a^{2}-45$ by $2\left(a-\frac{3}{2}\right)=2a-3$ to get $4a^{3}+6a^{2}+10a+15$. To factor the result, solve the equation where it equals to $0$.

$$4a^{3}+6a^{2}+10a+15=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $15$ and $q$ divides the leading coefficient $4$. List all candidates $\frac{p}{q}$.

$$±\frac{15}{4},±\frac{15}{2},±15,±\frac{5}{4},±\frac{5}{2},±5,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$a=-\frac{3}{2}$$

By Factor theorem, $a-k$ is a factor of the polynomial for each root $k$. Divide $4a^{3}+6a^{2}+10a+15$ by $2\left(a+\frac{3}{2}\right)=2a+3$ to get $2a^{2}+5$. To factor the result, solve the equation where it equals to $0$.

$$2a^{2}+5=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $2$ for $a$, $0$ for $b$, and $5$ for $c$ in the quadratic formula.

$$a=\frac{0±\sqrt{0^{2}-4\times 2\times 5}}{2\times 2}$$

Do the calculations.

$$a=\frac{0±\sqrt{-40}}{4}$$

Polynomial $2a^{2}+5$ is not factored since it does not have any rational roots.

$$2a^{2}+5$$

Rewrite the factored expression using the obtained roots.

$$\left(2a-3\right)\left(2a+3\right)\left(2a^{2}+5\right)$$

Evaluate

$8a^{4}+2a^{2}-45$